Question type

Vertex form questions on the SAT

In vertex form, $f (x) = a (x - h)^{2} + k$ , the vertex is $(h, k)$ . If $a$ is negative, $k$ is the maximum value; if positive, the minimum. To build the function, put the vertex in and use one more point to find $a$ . If two inputs give equal outputs, the vertex lies halfway between them.

These questions either give a quadratic in vertex form and ask for its maximum, minimum or another value, or give a vertex and a point and ask for the function.

It appeared on one of the five official practice tests we analysed, twice. Its Hard version hides the vertex: you have to find it from symmetry before you can write the form.

On 1 of 5 official practice tests
Desmos helps

Why students miss it

The sign of $h$ . In $- 2 (x + 4)^{2} + 18$ , the vertex is at $x = - 4$ , not 4.

Mixing up maximum and minimum. A negative $a$ opens downward, so $k$ is a maximum.

Missing the symmetry clue. $g (1) = g (9)$ means the vertex is at $x = 5$ , halfway between.

How to spot it

A squared bracket with something added: $a (x - h)^{2} + k$ .
"Maximum value", "minimum value" or "vertex" in the question.
Two inputs with the same output, such as $g (1) = g (9) = 12$ .

The method

Reading vertex form

Match the function to $a (x - h)^{2} + k$ .

$a < 0$: maximum value $k$ at $x = h$ .
$a > 0$: minimum value $k$ at $x = h$ .

Building it

Write $f (x) = a (x - h)^{2} + k$ with the vertex filled in.

One more point is given: substitute it to find $a$ .
Only two equal outputs: the vertex's $x$ is their midpoint.

With Desmos

Desmos marks the vertex of a parabola when you click it.

Type the function, for example y=-2(x+4)^2+18, and click the top of the curve: Desmos shows (−4, 18).
To build a function, type y=a(x-5)^2+20, add a slider for a, and move it until the curve passes through the given point.

Slider values are approximate; check the value of $a$ you found by substituting the point.

Two real questions

From the 1500prep bank, the way the app shows them. Answer first, then open the explanation.

1Question 1 of 2#724Adv

For the quadratic function $g$ , $g (1) = g (9) = 12$ , and the maximum value of $g$ is $20$ . Which equation could define $g$ ?

Pick an answer, then check it.

Show the answer and explanation

Equal outputs

g (1) = g (9)

sit symmetrically about the vertex, so the axis of symmetry is

x = \frac{1 + 9}{2} = 5

. A maximum means

a < 0

, and the vertex value is

20

g (x) = a (x - 5)^{2} + 20

. Use

g (1) = 12

a (1 - 5)^{2} + 20 = 12

, so

16 a = - 8

and

a = - \frac{1}{2}

2Question 2 of 2#723Adv

In the $x y$ -plane, the graph of the quadratic function $f$ has its vertex at $(- 2, 5)$ and passes through the point $(1, 32)$ . What is the value of $f (4)$ ?

Pick an answer, then check it.

Show the answer and explanation

Start from vertex form:

f (x) = a (x + 2)^{2} + 5

. Plug in

(1, 32)

9 a + 5 = 32

, so

a = 3

. Then

f (4) = 3 (4 + 2)^{2} + 5 = 3 (36) + 5 = 113

Questions students ask

How do you find the maximum value from vertex form?

In f(x) = a(x − h)² + k with a negative a, the maximum value is k, reached at x = h.

How do you find the vertex without vertex form?

If two x-values give the same output, the vertex's x-coordinate is halfway between them. You can also click the vertex on a Desmos graph.

Practise this type until it is automatic

Five more vertex-form questions, checked instantly, with every wrong choice explained. Free, no card.

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