Digital SAT Math · Advanced Math

Quadratic functions and graphs

Digital SAT Math · Advanced Math

A quadratic function can be written three ways, and the three ways are not stylistic preferences — each one puts a different fact about the parabola in plain sight and hides the other two. Standard form shows you where the curve crosses the $y$ -axis, factored form shows you where it crosses the $x$ -axis, and vertex form shows you the turning point and therefore the maximum or minimum. Almost every item on this skill is decided in the first five seconds, when you notice which of those three facts the question wants and rewrite toward it instead of grinding through the form you were handed.

On the test


Domain	Advanced Math (score report)
What it looks like	A parabola on a coordinate plane, or a quadratic printed in one of its three forms, or a real situation modelled by one
Often asked	“What is the minimum value of $f$ …?”, “Which equivalent form displays the $x$ -intercepts as constants…?”, “What is the vertex…?”, “Which equation defines the function shown…?”, “What is the best interpretation of the vertex…?”
Format	Multiple choice and student-produced response; the live digital sets on this subskill run roughly one grid-in in five
Calculator	Desmos reads a vertex off an ugly standard form instantly and is no help at all on the three archetypes that dominate here — see the box below

Recognition cues: the word parabola; a graph with one turning point; the phrases minimum value, maximum value, vertex, axis of symmetry, displays the … as constants or coefficients, for what value of $x$ does $f$ reach; and any model whose quantity rises and then falls — height, revenue, profit, area.

What this page owns

The College Board files quadratic and exponential behaviour together under “nonlinear functions”, and most of the web copies that heading onto one page. We split them, and we also split this page off from the equation-solving skill next door. Here the object is a function and the answer is a feature of its graph — a vertex, an intercept, an opening direction, a value at a point. When the answer is a value of $x$ that makes an equation true, that is nonlinear equations; when the rewrite is the whole answer and no graph is ever mentioned, that is equivalent expressions; when the curve bends but never turns around, that is exponential functions. The factoring is the same in all four places. What changes is what the factored form is for: over there it produces roots, here it produces the picture.

Pattern recognition

Six shapes cover nearly every item:

Read a form — the function is printed in vertex or factored form and the wanted feature is already sitting in it.
Choose a form — an equivalent rewrite is asked for, phrased as “displays the … as constants or coefficients”. Nothing is solved; you pick the form that exposes the feature.
Find the turning point — from standard form with $-\frac{b}{2a}$ , from factored form by averaging the roots, or by completing the square.
Graph to equation — a parabola is shown with its intercepts, and you rebuild the rule, leading coefficient included.
Equation to feature — a parabola is shown and one number is wanted: a minimum, a constant inside the rule, an $x$ where $f(x)=0$ .
Model in context — height, revenue, profit or area, where the vertex has units and the question asks either for its value or for where it occurs.

Which form for which question — the five-second decision. Wanted: the $x$ -intercepts, the zeros, the roots, where the graph crosses the $x$ -axis → factored form $a(x-r)(x-s)$ , and the intercepts are $r$ and $s$ with the printed signs reversed. Wanted: the vertex, the maximum, the minimum, the axis of symmetry → vertex form $a(x-h)^2+k$ , and the turning point is $(h,k)$ with the sign of $h$ reversed. Wanted: the $y$ -intercept, or the value at $x = 0$ → standard form $ax^2+bx+c$ , and the answer is $c$ with no work at all. Wanted: which way it opens, or whether the turning point is a max or a min → the sign of $a$ , which is visible in all three forms. The rewrite is the work. Once you are in the right form the answer is a constant you copy, which is why an item that looks like thirty seconds of algebra is usually five seconds of reading plus one line.

Method

Name the feature the question wants, in the question’s own words: an intercept, a turning point, a value, a location, an equation. Do this before you look at the coefficients — the coefficients will suggest work you do not need.
Check the form you were handed against the feature you want. If they match, the answer is a constant in the expression and you are done. If they do not, rewrite: factor, complete the square, or expand.
Reverse the sign inside every parenthesis. $(x - h)$ means the graph moved $h$ units to the right, so $f(x) = 2(x+3)^2 - 5$ has its vertex at $x = -3$ , not $x = 3$ . The same reversal turns the factor $(x+7)$ into the intercept $-7$ . This one rule is behind more wrong answers on this skill than all the arithmetic combined.
Do not reverse the sign of anything outside a parenthesis. In $a(x-h)^2 + k$ , the $k$ is the vertex’s $y$ -coordinate exactly as printed. In $a(x-r)(x-s)$ , the $a$ is the leading coefficient exactly as printed.
Separate the value from the location. The maximum is a $y$ ; the input where it happens is an $x$ . $-\frac{b}{2a}$ gives you the $x$ , and you are not finished — substitute it back to get the $y$ . Both numbers will be among the choices.
Answer in the units of the question. In a model, the vertex is a pair of measured quantities: dollars and dollars-per-item, feet and seconds. Reading “at what price” and reporting the revenue is the same error as step 5, wearing a costume.

Printed form	Displays immediately	Costs one step	Costs real work
$ax^2+bx+c$	$y$ -intercept $c$ ; opening direction	vertex $x$ via $-\frac{b}{2a}$	intercepts (factor or formula)
$a(x-r)(x-s)$	$x$ -intercepts $r$ and $s$ ; opening direction	vertex $x$ via $\frac{r+s}{2}$	$y$ -intercept (multiply out, or evaluate at 0)
$a(x-h)^2+k$	vertex $(h,k)$ ; axis $x=h$ ; max or min value $k$	$y$ -intercept (evaluate at 0)	$x$ -intercepts (solve for $x$ )

Completing the square, and the one place it goes wrong. To move from $ax^2+bx+c$ to vertex form, factor $a$ out of the first two terms only, halve the coefficient of $x$ that remains, square it, and — this is the step — add and subtract that square inside the parentheses, then carry it out past the $a$ . For $f(x)=3x^2+12x+5$ : factor to $3(x^2+4x)+5$ , halve 4 to get 2, square it to get 4, and write $3\big[(x+2)^2-4\big]+5 = 3(x+2)^2-12+5 = 3(x+2)^2-7$ . The vertex is $(-2,-7)$ and the minimum value is $-7$ . The error is subtracting the 4 instead of the 12: the completed square was created inside a bracket that gets multiplied by 3, so the correction leaving the bracket is $3 \times 4$ . Check by expanding backwards every time — $3(x+2)^2-7 = 3x^2+12x+12-7 = 3x^2+12x+5$ ✓ — and note that when $a=1$ this whole hazard disappears, which is why practice on $a=1$ does not prepare you for the item that ships $a=3$ .

Reading a parabola you are handed. A graph gives you more than it looks like. Where it crosses the $x$ -axis gives you the factors — crossings at $-1$ and $3$ mean factors $(x+1)$ and $(x-3)$ , signs reversed again. Where it crosses the $y$ -axis gives you $f(0)$ , and that is the number that pins down the leading coefficient, which the intercepts alone can never tell you: infinitely many parabolas share a pair of $x$ -intercepts, and they differ only in $a$ . So the full recovery is three moves — write $f(x)=a(x-r)(x-s)$ from the crossings, substitute the third point, solve for $a$ . Skipping the third move produces an equation with the right shape and the wrong width, and that equation is always one of the choices. Two free readings come with the picture: the turning point sits exactly halfway between the crossings, and the parabola opens upward exactly when $a>0$ , which you can see before you compute anything.

Desmos on this skill — the one thing it is great at, and the three it cannot touch. Type a quadratic in $y=$ form and Desmos plots it; click the curve and it labels the vertex, the zeros and the $y$ -intercept as grey points you can read straight off. That makes it genuinely the fastest route to a turning point from an ugly standard form — $y = 4.2x^2 - 19.6x + 11$ is a twenty-second job by hand and a five-second job on screen, with no chance of a $-\frac{b}{2a}$ sign slip. Where it earns you nothing: a “which form displays” item, where every choice is an equivalent expression and the graphs are identical, so plotting all four tells you only that they coincide — the question is about what a form shows, which is a reading question, not a graphing one. A letter in place of a coefficient has nothing to plot until you decide the letter’s value; a slider gets you near the answer and will not give you the exact number a grid-in wants. And a context item still needs you to decide whether the answer is the vertex’s $x$ or its $y$ , which Desmos will happily show you both of. One habit worth keeping: when Desmos reports a vertex at $(2.5,\,-6.25)$ , the grid-in accepts $-25/4$ , and an exact fraction is always safer than a rounded decimal.

Worked example 1 — one function, three forms, three questions

Stem. The function $f$ is defined by $f(x) = x^2 - 6x + 8$ . (a) What is the $y$ -intercept of the graph of $y=f(x)$ ? (b) What are its $x$ -intercepts? (c) What is the minimum value of $f$ ?

Step 1 — part (a) needs no work, because the printed form already displays it. In standard form the constant term is $f(0)$ : $f(0) = 0 - 0 + 8 = 8$ , so the $y$ -intercept is $(0, 8)$ .

Step 2 — part (b) wants $x$ -intercepts, so rewrite into factored form. Two integers multiplying to 8 and adding to $-6$ are $-2$ and $-4$ :

f(x) = (x-2)(x-4)

A product is zero exactly when a factor is zero, so the graph crosses at $x=2$ and $x=4$ . Note the reversal: the factors carry minus signs and the intercepts are positive.

Step 3 — part (c) wants a minimum, so rewrite into vertex form. Halve $-6$ to get $-3$ , square it to get 9, and add and subtract:

f(x) = x^2 - 6x + 9 - 9 + 8 = (x-3)^2 - 1

The vertex is $(3,-1)$ . Since $a = 1 > 0$ the parabola opens upward, so $-1$ is a minimum, and the minimum value is $-1$ — the number $3$ is where the minimum happens, not the minimum.

Check. Expand both rewrites back: $(x-2)(x-4) = x^2-6x+8$ ✓ and $(x-3)^2-1 = x^2-6x+9-1 = x^2-6x+8$ ✓. Cross-check the vertex a second way: it must sit halfway between the intercepts, and $\frac{2+4}{2}=3$ ✓, with $f(3)=9-18+8=-1$ ✓. Answers: (a) $(0,8)$ ; (b) $x=2$ and $x=4$ ; (c) $-1$ .

Trap watch. Reporting $3$ for part (c) answers the wrong one of the vertex’s two coordinates. Reporting $-2$ and $-4$ for part (b) copies the constants out of the factors without reversing their signs. And rewriting is only ever legal if the result expands back to what you started with — a single expansion check kills every arithmetic slip in this example.

Worked example 2 — factored form, a downward parabola, and the value-versus-location split

Stem. The function $g$ is defined by $g(x) = -2(x+1)(x-5)$ . At what value of $x$ does $g$ reach its maximum, and what is that maximum value?

Step 1 — read the intercepts, reversing signs. The factors $(x+1)$ and $(x-5)$ vanish at $x=-1$ and $x=5$ , so the graph crosses the $x$ -axis there.

Step 2 — the axis of symmetry is halfway between the crossings. A parabola is symmetric about the vertical line through its vertex, and the two crossings are mirror images, so

x = \frac{-1+5}{2} = 2

Step 3 — the question also wants the value, so substitute. The location is $x=2$ ; the value is $g(2)$ :

g(2) = -2(2+1)(2-5) = -2(3)(-3) = 18

Step 4 — confirm it is a maximum, not a minimum. The leading coefficient is $-2 < 0$ , so the parabola opens downward and the turning point is the highest point on the curve.

Check. Expand and use the other route: $g(x) = -2(x^2-4x-5) = -2x^2+8x+10$ , so $-\frac{b}{2a} = -\frac{8}{2(-2)} = 2$ ✓ and $g(2) = -8+16+10 = 18$ ✓. Symmetry test: $g(0) = 10$ and $g(4) = -2(5)(-1) = 10$ — equal outputs at inputs equidistant from 2, as they must be ✓. Answer: the maximum value $18$ occurs at $x = 2$ .

Trap watch. “18” and “2” are both correct numbers answering different questions, and both will be offered. Averaging the printed constants instead of the roots gives $\frac{1+(-5)}{2} = -2$ , the sign-reversal error in its most disguised form. And forgetting to check the sign of $a$ leaves you calling 18 a minimum on a curve that has no minimum at all.

Worked example 3 — from a graph back to the equation

Stem. A parabola in the $xy$ -plane crosses the $x$ -axis at $(-2,0)$ and $(6,0)$ and passes through $(0,-36)$ . Which equation defines the function $f$ whose graph this is, and what is the minimum value of $f$ ?

Step 1 — the crossings give the factors, with signs reversed. Crossings at $-2$ and $6$ mean the factors are $(x+2)$ and $(x-6)$ , so

f(x) = a(x+2)(x-6)

for some constant $a$ that the crossings alone cannot determine — every parabola with those two intercepts fits this template, wide or narrow, up or down.

Step 2 — the third point pins down $a$ . Substitute $(0,-36)$ :

-36 = a(0+2)(0-6) = -12a \qquad \Rightarrow \qquad a = 3

so $f(x) = 3(x+2)(x-6)$ .

Step 3 — the minimum needs the vertex, and the crossings hand you its location. The axis of symmetry is halfway between $-2$ and $6$ :

x = \frac{-2+6}{2} = 2, \qquad f(2) = 3(4)(-4) = -48

and since $a = 3 > 0$ , $-48$ is a minimum.

Check. Expand: $f(x) = 3(x^2-4x-12) = 3x^2-12x-36$ . The $y$ -intercept is $-36$ ✓, matching the given point; $-\frac{b}{2a} = \frac{12}{6} = 2$ ✓; $f(2) = 12-24-36 = -48$ ✓; and $f(-2) = 12+24-36 = 0$ ✓. Answers: $f(x) = 3(x+2)(x-6)$ , minimum value $-48$ .

Trap watch. Writing $f(x) = (x+2)(x-6)$ and stopping gives a parabola through both crossings whose $y$ -intercept is $-12$ , not $-36$ — right shape, wrong width, and it is always a choice. Writing $(x-2)(x+6)$ reverses the signs in the wrong direction and produces crossings at $2$ and $-6$ . And reporting $2$ as the minimum confuses where the minimum happens with what it is, for the third time on this page — which is deliberate, because it is the single most reliable way to lose a point on this skill.

Practice

Answer before opening the explanation. Two items are student-produced response, matching the grid-in share of the live digital sets on this subskill; three hand you a parabola and no equation, because on this skill the graph is the question more often than it is an illustration. Every wrong choice below is one named misconception — a sign not reversed inside a parenthesis, a leading coefficient dropped, the vertex’s $x$ reported where its $y$ was asked for, a completed square not multiplied back out. When you miss one, log the misconception, not the item number.

12 questions — 10 multiple choice, 2 student-produced response. Every wrong choice has its own explanation.

Question 1 Warm-up

The function f is defined by f(x) = 2(x + 3)² - 5. What is the vertex of the graph of y = f(x) in the xy-plane?

Show the answer Choice A

Why it is right

Vertex form is f(x) = a(x - h)² + k, and the vertex is the point (h, k). Matching 2(x + 3)² - 5 against that template literally, the expression x + 3 is x - (-3), so h = -3, and the constant added outside the squared term is k = -5. The vertex is therefore (-3, -5). Confirm it independently by expanding: f(x) = 2x² + 12x + 18 - 5 = 2x² + 12x + 13, so -b/(2a) = -12/4 = -3 and f(-3) = 18 - 36 + 13 = -5.

Why each other choice fails

Choice B: Copies the 3 out of (x + 3) without reversing its sign. The template subtracts h, so a plus sign inside the parentheses means a negative h; at x = 3 the function value is 2(36) - 5 = 67, nowhere near the turning point.
Choice C: Reverses the sign of the constant that sits outside the squared term as well. Only the number grouped with x inside the parentheses flips; the k in a(x - h)² + k is the vertex height exactly as printed, so it stays -5.
Choice D: Reads the leading coefficient 2 as the vertex's x-coordinate. The 2 controls how narrow the parabola is and that it opens upward; it plays no part in locating the turning point, which is fixed by the 3 inside the parentheses.

Question 2 Standard

The function f is defined by f(x) = (x - 6)(x + 2). At which points does the graph of y = f(x) intersect the x-axis?

Show the answer Choice C

Why it is right

The graph meets the x-axis where f(x) = 0, and a product is zero exactly when one of its factors is zero. So x - 6 = 0 gives x = 6, and x + 2 = 0 gives x = -2. The intersection points are (6, 0) and (-2, 0). Substituting confirms both: f(6) = (0)(8) = 0 and f(-2) = (-8)(0) = 0. Note that the constants printed inside the factors are -6 and +2 while the intercepts are +6 and -2 — each sign reverses.

Why each other choice fails

Choice A: Has the right numbers in the wrong slots. A point where the graph crosses the x-axis has y = 0, so the zero belongs in the second coordinate; (0, 6) and (0, -2) are points on the y-axis, and this graph meets the y-axis only once, at (0, -12).
Choice B: Copies the constants out of the factors without reversing either sign. Testing them shows why that fails: f(-6) = (-12)(-4) = 48 and f(2) = (-4)(4) = -16, neither of which is zero.
Choice D: Reverses the sign of the first factor's constant but copies the second one as printed. The rule applies to every factor: x + 2 = 0 gives x = -2, and f(2) = (-4)(4) = -16 confirms that 2 is not a zero.

Question 3 Standard

The function f is defined by f(x) = x² - 2x - 24. Which of the following equivalent forms of f(x) displays the x-intercepts of the graph of y = f(x) as constants or coefficients?

Show the answer Choice B

Why it is right

An x-intercept is an input that makes the output zero, and a product is zero exactly when a factor is zero, so the form that displays the intercepts is the fully factored one. Two integers multiplying to -24 and adding to -2 are -6 and +4, giving (x - 6)(x + 4), which expands back to x² + 4x - 6x - 24 = x² - 2x - 24 as required. Its factors vanish at x = 6 and x = -4, so those constants are the intercepts, sign-reversed as always.

Why each other choice fails

Choice A: This is a genuine equivalent form — (x - 1)² - 25 expands to x² - 2x + 1 - 25 = x² - 2x - 24 — but it is vertex form, so what it displays is the turning point (1, -25) and the minimum value -25, not the intercepts. It is the right answer to a different question.
Choice C: Also equivalent, since x(x - 2) - 24 = x² - 2x - 24, but it is not a product set against zero: the -24 sits outside, so x = 0 and x = 2 are not zeros of f. Checking, f(0) = -24 and f(2) = -24, which is exactly the value this form does display, the y-intercept.
Choice D: Not equivalent to f at all. Expanding gives x² - 4x + 6x - 24 = x² + 2x - 24, whose middle term has the wrong sign, so this is a different function with intercepts at -6 and 4. Reversing the signs inside both factors changes the product's linear term.

Question 4 Standard

The function f is defined by f(x) = 3x² - 12x + 5. What is the minimum value of f?

Show the answer Choice D

Why it is right

The leading coefficient 3 is positive, so the parabola opens upward and its lowest point is the vertex. The vertex's x-coordinate is -b/(2a) = -(-12)/(2·3) = 12/6 = 2, and the minimum value is the function evaluated there: f(2) = 3(4) - 24 + 5 = 12 - 24 + 5 = -7. Completing the square confirms it independently: 3(x² - 4x) + 5 = 3[(x - 2)² - 4] + 5 = 3(x - 2)² - 12 + 5 = 3(x - 2)² - 7, whose constant term is the minimum value -7.

Why each other choice fails

Choice A: Uses x = -2 instead of x = 2, the sign slip in -b/(2a) when b is already negative, and then evaluates: f(-2) = 3(4) + 24 + 5 = 41. Since 41 is larger than f(0) = 5, it cannot be a minimum of an upward-opening parabola.
Choice B: Stops at -b/(2a) = 2 and reports it. That number is the input at which the minimum occurs, not the minimum itself; the question asks for a value of f, so 2 still has to be substituted back into the function.
Choice C: Takes the constant term of the standard form as the minimum. In f(x) = ax² + bx + c the constant c is the y-intercept, f(0) = 5, and since the vertex sits at x = 2 rather than x = 0, f(0) is not the lowest value the function reaches.

Question 5 Standard

The graph of y = f(x) is shown in the xy-plane, where f is a quadratic function. What is the minimum value of f?

Figure 1. The graph of y = f(x), with its x-intercepts, y-intercept and vertex marked.

Show the answer Choice B

Why it is right

The parabola opens upward, so its lowest point is the marked vertex (2, -9). The minimum value of a function is an output, so it is the y-coordinate of that lowest point: -9. Everything on the graph agrees — the curve is symmetric about the vertical line through x = 2, the two marked x-intercepts -1 and 5 sit three units either side of it, and no plotted point lies below y = -9.

Why each other choice fails

Choice A: Reports the x-coordinate of the vertex, which is where the minimum occurs rather than what it is. The value 2 is an input; substituting it gives the output -9, which is what the question asked for.
Choice C: Reads the y-intercept (0, -5) as the lowest point. The graph continues to fall to the right of the y-axis and does not turn around until x = 2, so -5 is a value the function takes on the way down, not its smallest.
Choice D: Takes the x-intercept at (-1, 0) and reports its x-coordinate. That point is where the curve crosses the axis, not where it turns; -1 is neither the location nor the value of the minimum.

Question 6 Harder

The function g is defined by g(x) = (x + 7)(x - 3). What is the x-coordinate of the vertex of the graph of y = g(x) in the xy-plane?

Show the answer Choice A

Why it is right

The factors vanish at x = -7 and x = 3, so those are the x-intercepts. A parabola is symmetric about the vertical line through its vertex, and the two intercepts are mirror images across that line, so the vertex sits at their average: (-7 + 3)/2 = -4/2 = -2. Expanding gives an independent check: g(x) = x² + 4x - 21, so -b/(2a) = -4/2 = -2, and the symmetry holds numerically since g(-3) = (4)(-6) = -24 and g(-1) = (6)(-4) = -24.

Why each other choice fails

Choice B: Averages the constants exactly as they are printed inside the factors, (7 + (-3))/2 = 2, instead of averaging the intercepts. Each constant has to be sign-reversed first: (x + 7) gives the intercept -7, not 7.
Choice C: Adds the two intercepts, -7 + 3 = -4, and forgets to halve. The axis of symmetry is the midpoint of the intercepts, so the sum must be divided by 2; -4 is twice as far from the vertex as it should be.
Choice D: Multiplies the intercepts instead of averaging them, giving (-7)(3) = -21. That product is the y-intercept, g(0) = -21, which is a value of the function rather than a location on the x-axis.

Question 7 Harder

The graph of y = f(x) is shown in the xy-plane, where f is a quadratic function. Which equation defines f?

Figure 2. The graph of y = f(x), with both x-intercepts and the y-intercept marked.

Show the answer Choice D

Why it is right

The curve crosses the x-axis at -1 and 3, so the factors are (x + 1) and (x - 3) with the signs reversed, and f(x) = a(x + 1)(x - 3) for some constant a. The intercepts alone cannot fix a, so use the third marked point: f(0) = -6 gives a(1)(-3) = -6, hence -3a = -6 and a = 2. So f(x) = 2(x + 1)(x - 3), which expands to 2x² - 4x - 6. All three marked points check: f(-1) = 0, f(3) = 0, and f(0) = -6.

Why each other choice fails

Choice A: Has the right factors and the right magnitude but the wrong sign of a, so it opens downward while the graph shown opens upward. It also fails the y-intercept: -2(1)(-3) = 6, not -6.
Choice B: Copies the intercepts into the factors without reversing their signs, so its zeros are 1 and -3 instead of -1 and 3. Testing the graph's own crossing point kills it: this rule gives 2(-2)(2) = -8 at x = -1, not 0.
Choice C: Has the right factors but drops the leading coefficient, which the two x-intercepts can never determine — infinitely many parabolas share a pair of intercepts and differ only in a. This one passes through (0, -3) rather than the graph's y-intercept (0, -6), so it is the correct shape at half the vertical scale.

Question 8 Harder Student-produced response

A model rocket is launched upward from a platform. Its height h, in feet above the ground, t seconds after launch is modeled by h(t) = -16t² + 96t + 40. What is the maximum height, in feet, that the rocket reaches?

Your answer

Show the answer 184

Why it is right

The leading coefficient -16 is negative, so the parabola opens downward and the greatest height is the vertex value. The vertex occurs at t = -b/(2a) = -96/(2 · -16) = -96/-32 = 3 seconds, and the height there is h(3) = -16(9) + 96(3) + 40 = -144 + 288 + 40 = 184 feet. Vertex form confirms it: -16(t² - 6t) + 40 = -16[(t - 3)² - 9] + 40 = -16(t - 3)² + 144 + 40 = -16(t - 3)² + 184.

Answers students type instead

3: This is the time at which the maximum happens, not the height. The vertex has two coordinates and the question asks for the one measured in feet, so 3 seconds still has to be substituted back into h.
40: This is h(0), the platform height at the instant of launch. It is where the rocket starts, not the highest point it reaches; the rocket climbs for another 144 feet before turning around.
472: Comes from evaluating with +16t² instead of -16t², giving 144 + 288 + 40 = 472. With a positive leading coefficient the parabola would open upward and have no maximum at all, so this answer contradicts the model it came from.

Question 9 Harder

The graph of the quadratic function f in the xy-plane passes through the points (2, 5) and (10, 5). What is the x-coordinate of the vertex of the graph?

Show the answer Choice C

Why it is right

A parabola is symmetric about the vertical line through its vertex, so two inputs with the same output must be the same distance from that line on opposite sides. The axis of symmetry therefore sits at the average of the inputs: (2 + 10)/2 = 12/2 = 6. No equation is needed. As a sanity check, any quadratic through those points has the form f(x) = a(x - 2)(x - 10) + 5, whose x-intercept midpoint is again (2 + 10)/2 = 6.

Why each other choice fails

Choice A: Reports 5, the shared output value. That number is a y-coordinate — the height at which the parabola cuts across those two points — while the question asks for an x-coordinate.
Choice B: Halves the distance between the two inputs, (10 - 2)/2 = 4, instead of averaging them. The value 4 is how far the vertex lies from each point, not where it is; adding it to the smaller input, 2 + 4, recovers the correct answer 6.
Choice D: Adds the two inputs, 2 + 10 = 12, and forgets to divide by 2. A midpoint is a sum halved; 12 lies outside the interval between the given points entirely, so it cannot be equidistant from them.

Question 10 Hardest

The function f is defined by f(x) = 2x² - 12x + 23. The function can be written in the form f(x) = a(x - h)² + k, where a, h and k are constants. What is the value of k?

Show the answer Choice B

Why it is right

Factor the leading coefficient out of the first two terms only: 2(x² - 6x) + 23. Half of -6 is -3 and (-3)² = 9, so complete the square inside the bracket and correct for it outside: 2[(x - 3)² - 9] + 23 = 2(x - 3)² - 18 + 23 = 2(x - 3)² + 5. So k = 5. Two independent checks: expanding gives 2(x² - 6x + 9) + 5 = 2x² - 12x + 18 + 5 = 2x² - 12x + 23, and the vertex route gives -b/(2a) = 12/4 = 3 with f(3) = 18 - 36 + 23 = 5.

Why each other choice fails

Choice A: Reports h rather than k. The value 3 is the vertex's x-coordinate — where the minimum occurs — while k is the constant added outside the squared term, the vertex's height.
Choice C: Copies the constant term of the standard form. That number is f(0) = 23, the y-intercept, and it equals k only when the vertex happens to sit on the y-axis, which here it does not.
Choice D: Subtracts 9 instead of 18: 23 - 9 = 14. The completed square was created inside a bracket that the 2 multiplies, so the amount that has to come back out is 2 · 9 = 18. Expanding 2(x - 3)² + 14 gives 2x² - 12x + 32, not the original function.

Question 11 Hardest Student-produced response

The function f is defined by f(x) = a(x - 4)² + 11, where a is a constant. The graph of y = f(x) in the xy-plane passes through the point (6, 3). What is the value of a?

Your answer

Show the answer -2

Why it is right

The point (6, 3) lies on the graph, so f(6) = 3. Substituting gives a(6 - 4)² + 11 = 3, that is a(2)² + 11 = 3, so 4a + 11 = 3 and 4a = -8, giving a = -2. Check by rebuilding the function: f(x) = -2(x - 4)² + 11 gives f(6) = -2(4) + 11 = -8 + 11 = 3 as required, and the vertex is (4, 11), a maximum since a is negative.

Answers students type instead

2: Solves 4a = 8 instead of 4a = -8, dropping the sign when moving 11 to the other side. The sign matters: with a = 2 the graph would pass through (6, 19), and the parabola would open upward with 11 as a minimum rather than a maximum.
-8: Stops at the intermediate line 4a = -8 and reports that. The equation still has to be divided by 4, since a multiplies the squared factor (6 - 4)² = 4 rather than standing alone.
-4: Uses 6 - 4 = 2 without squaring it, giving 2a + 11 = 3 and a = -4. The template is a(x - h)², so the entire difference is squared before a multiplies it.

Question 12 Hardest

A community pool models its daily revenue R, in dollars, as a function of the price p, in dollars, that it charges for a day pass. The graph of the model is shown. Which of the following is the best interpretation of the vertex of the graph?

Figure 3. Modeled daily revenue against day-pass price, with the two zero-revenue prices and the peak marked.

Show the answer Choice D

Why it is right

The vertex is the marked highest point of the curve, (10, 500). Its coordinates carry the units of the axes: the input p is a price in dollars per day pass and the output R is revenue in dollars, so the vertex says that revenue peaks at $500 and that this peak happens when the price is set at $10. The parabola opens downward, which is what makes the turning point a maximum rather than a minimum.

Why each other choice fails

Choice A: Reads the vertex's height as a rate of change, which is what a slope would mean on a straight-line model. This graph is not a line — revenue rises up to p = 10 and falls after it — so no single number describes the change per additional dollar of price.
Choice B: Swaps the two coordinates of the vertex. The first coordinate is measured along the horizontal axis, which is the price, and the second along the vertical axis, which is the revenue; a $500 day pass sits far off the right edge of this model, where revenue has long since fallen to zero.
Choice C: Treats one point on the curve as if it described the whole interval between the two zeros. The graph reaches $500 at exactly one price; at every other price in that range the revenue is strictly less, falling to $0 at both p = 0 and p = 20.

Common mistakes

Not reversing the sign inside a parenthesis — $2(x+3)^2-5$ has its vertex at $x=-3$ ; the factor $(x+7)$ gives the intercept $-7$ . Match against $a(x-h)^2+k$ literally: $x+3$ is $x-(-3)$ .
Reversing the sign of something outside a parenthesis — the $k$ in $a(x-h)^2+k$ is the vertex’s height exactly as printed. Only the number sitting with $x$ inside the parentheses flips.
Reporting where instead of what — a “minimum value” is a $y$ , and $-\frac{b}{2a}$ is an $x$ . Substitute back. Both numbers are in the choices, every time.
Dropping the leading coefficient when rebuilding from a graph — the $x$ -intercepts fix the factors and nothing else; only a third point fixes $a$ . Check your equation against the $y$ -intercept before choosing.
Subtracting the wrong amount when completing the square — with $a \ne 1$ , the square is completed inside a bracket that is multiplied by $a$ , so the correction that leaves the bracket is $a$ times as big. Expand backwards to confirm.
Confusing the vertex with an intercept — the turning point is on the $x$ -axis only when the parabola touches it once. In general the vertex is not a zero and a zero is not the vertex.
Assuming every parabola opens upward — the sign of $a$ decides, and a downward parabola has a maximum and no minimum at all. “Minimum value” on a curve with $a<0$ is a question with no answer.
Assuming two $x$ -intercepts — a parabola can cross twice, touch once, or miss the $x$ -axis entirely. If factoring will not work, that may be the reason rather than your arithmetic.
Averaging the printed constants instead of the roots — for $(x+7)(x-3)$ the axis is at $\frac{-7+3}{2} = -2$ , not $\frac{7-3}{2} = 2$ . Reverse the signs first, then average.
Writing the axis of symmetry as $y = h$ — it is a vertical line, so it is $x = h$ . The same slip turns a correct vertex into a wrong coordinate pair.
Computing when the question said “displays” — an item asking which equivalent form shows a feature is answered by reading, not by solving. Every choice that is genuinely equivalent gives the same graph; you are choosing what is visible, and the fastest disqualification is expanding a choice and finding it is not equivalent at all.
Answering in the wrong units in a model — “at what price is revenue greatest” wants dollars per item, “what is the greatest revenue” wants dollars. The vertex supplies both; the sentence decides which.

FAQ

Do I need to memorize $x = -\frac{b}{2a}$ ? Yes, and it is worth the ten seconds: it is not on the Digital SAT reference sheet, it works on any standard form without rewriting, and it is faster than completing the square when all you want is the vertex. Remember it gives you the $x$ -coordinate only — the value of the function there still has to be computed.

Vertex form or $-\frac{b}{2a}$ ? If the question wants only the turning point, use $-\frac{b}{2a}$ and substitute; it is two short lines. If the question wants the function written in vertex form — because a choice must display the minimum as a constant — you have to complete the square, since $-\frac{b}{2a}$ produces numbers, not an expression.

How do I find the vertex if the quadratic will not factor? Nothing about the vertex requires factoring. $-\frac{b}{2a}$ needs only the coefficients, and completing the square works on any quadratic at all. Factoring is for intercepts, and a parabola that will not factor over the integers may have irrational intercepts or none.

What exactly is an item asking when it says “displays the $x$ -intercepts as constants or coefficients”? It is asking which rewrite makes the intercepts literally readable as numbers in the expression. All the choices are usually equivalent, so none is “more correct” as a function — you are being tested on what each form reveals. Answer it by reading, and use the check in reverse: if a choice is not equivalent to the original, it is wrong for a second, blunter reason.

Is the axis of symmetry the same as the vertex? No. The axis is a vertical line, $x = h$ ; the vertex is the point $(h,k)$ where that line meets the curve. An item asking for the axis wants an equation of a line, and an item asking for the vertex wants an ordered pair.

A parabola passes through two points with the same $y$ . Can I use that? It is one of the fastest facts on this skill. Equal outputs mean equal distances from the axis of symmetry, so the axis sits at the average of the two inputs — no equation needed, and it works even when the function is never given to you.

How many of these show up on a test? Advanced Math is roughly a third of Digital SAT Math, and quadratic and exponential functions together are the largest named piece of it; free guides put nonlinear functions at several questions across the two modules. The equation-solving half lives on its own page; what this page drills is the function-and-graph half.

Grid-in format? Enter the exact value. Negative answers are allowed, fractions are accepted as fractions, and the most common grid-in loss on this skill is entering the vertex’s $x$ when the item asked for its $y$ — there are no answer choices to warn you that you have picked the wrong coordinate.

Coming next in this domain: interpreting nonlinear graphs.

Editorial team · last reviewed 2026-08-06

Quadratic functions and graphs

On the test

What this page owns

Pattern recognition

Method

Worked example 1 — one function, three forms, three questions

Worked example 2 — factored form, a downward parabola, and the value-versus-location split

Worked example 3 — from a graph back to the equation

Practice

Question 1 Warm-up

Why it is right

Why each other choice fails

Question 2 Standard

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Question 3 Standard

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Question 4 Standard

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Question 5 Standard

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Question 6 Harder

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Question 7 Harder

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Question 8 Harder Student-produced response

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Answers students type instead

Question 9 Harder

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Question 10 Hardest

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Question 11 Hardest Student-produced response

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Question 12 Hardest

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Common mistakes

FAQ

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