SAT Math · Advanced Math · free lesson

Vertex Form and Graphing Parabolas

The idea, then worked examples solved step by step, then practice questions whose numbers change every time you try them.

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Vertex Form and Graphing Parabolas

The vertex of $y = x^2 - 6x + 5$ is the lowest point.

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Read what is said

The graph of a quadratic is a parabola. It comes down, turns once, and goes back up. That turning point is the vertex, and the SAT asks about it constantly: the lowest or highest value, and where it happens.
Vertex form shows the vertex without any work. A square is never negative, so the squared part is smallest when x is three, and that leaves negative four as the least value of y.
Here is the trap. The form is x minus h, so the sign inside the parentheses flips. This vertex is at x equal to negative two, not two. The number outside keeps its sign.
Now an example in standard form. We want the x where this function is least, and also the minimum value. Those are two different numbers, so read the question carefully.
Here a is one and b is negative ten, so negative b is ten. The x of the vertex is negative b divided by two a.
Now put five back into the function to get the y of the vertex. The minimum is negative eighteen, reached at x equal to five. The vertex form confirms both numbers at once.
Remember this. If the question asks for what value of x, the answer is the x of the vertex, five. If it asks for the minimum value, the answer is the y, negative eighteen. Now practice it with new numbers.

The idea

The graph of a quadratic

The graph of a quadratic function such as $y = x^2 - 6x + 5$ is a curve called a parabola. It comes down, turns once, and goes back up. The turning point is the vertex. Here it is the lowest point of the graph, so its $y$ is the minimum value of the function.

The dashed line through the vertex is the axis of symmetry. The left half of the parabola is the mirror image of the right half. So two points at the same height are always the same distance from the axis: the zeros 1 and 5 are each 2 away from 3.

Rule.

In $y = ax^{2} + bx + c$ , the sign of $a$ decides the direction. If $a$ is positive the parabola opens up and the vertex is the minimum. If $a$ is negative it opens down and the vertex is the maximum.

Three forms, one parabola

The same quadratic can be written in three ways. The engine multiplies out the second and third, and each comes back to the first:

\begin{aligned} y &= (x - 3)^{2} - 4 \\ &= x^{2} - 6x + 5 \\ y &= (x - 1)(x - 5) \\ &= x^{2} - 6x + 5 \end{aligned}

Each form shows something different without any work.

Form	This parabola	What you read from it
Standard: $y = ax^{2} + bx + c$	$y = x^2 - 6x + 5$	the $y$ -intercept $c$ , here 5
Factored: $y = a(x - p)(x - q)$	$y = (x - 1)(x - 5)$	the zeros $p$ and $q$ , here 1 and 5
Vertex: $y = a(x - h)^{2} + k$	$y = (x - 3)^{2} - 4$	the vertex $(h, k)$ , here $(3, -4)$

Why does the vertex form show the vertex? A square is never negative, so $(x - 3)^{2}$ is at least 0, and it is 0 only when $x$ is 3. That makes $-4$ the least value $y$ can have, reached at $x = 3$ .

Warning.

Watch the sign inside the parentheses. The form is $(x - h)^{2}$ , so $y = (x + 2)^{2} + 7$ has its vertex at $x = -2$ , not at 2. The number outside keeps its sign: the vertex is $(-2, 7)$ .

The vertex from the other two forms

From the factored form, use the symmetry. The vertex is halfway between the zeros. From the standard form, the $x$ of the vertex is $-\frac{b}{2a}$ ; here $b$ is $-6$ , so $-b$ is 6. Either way, put that $x$ back into the function to get the $y$ of the vertex.

\begin{aligned} h &= \frac{1 + 5}{2} \\ &= 3 \\ h &= \frac{6}{2 \times 1} \\ &= 3 \\ k &= h^{2} - 6h + 5 \\ &= 3^{2} - 6 \times 3 + 5 \\ &= -4 \end{aligned}

Both give $x$ $=$ $3$ , and the function is $-4$ there, as the vertex form said.

Move the parabola

Drag the sliders of $y = a(x - h)^{2} + k$ . Changing $h$ slides the parabola left and right, and $k$ slides it up and down: the red vertex is always at $(h, k)$ . Changing $a$ makes the parabola narrow or wide, and a negative $a$ turns it over.

A ball in the air

A ball is thrown up from a roof. Its height in feet after $t$ seconds is $-16t^2 + 64t + 80$ . The coefficient of $t^{2}$ is negative, so the parabola opens down and the vertex is the highest point of the flight. Watch the height rise, stop for an instant, and fall.

The three forms answer three questions about the flight:

\begin{aligned} y &= -16(t - 2)^{2} + 144 \\ &= -16t^{2} + 64t + 80 \\ y &= -16(t - 5)(t + 1) \\ &= -16t^{2} + 64t + 80 \end{aligned}

Standard form, $-16t^2 + 64t + 80$ : the ball starts 80 feet up (the value at $t = 0$ ).
Vertex form, $-16(t - 2)^2 + 144$ : the greatest height is 144 feet, reached after 2 seconds.
Factored form, $-16(t - 5)(t + 1)$ : the height is 0 at $t = 5$ , when the ball lands. The other zero, $-1$ , is before the throw and has no meaning here.

Note.

On the SAT, read what is asked before you answer. "At what value of $x$ is the function least?" wants the $x$ of the vertex. "What is the minimum value?" wants the $y$ . Both numbers are usually among the choices.

Worked examples

Each example is one of the kinds of question the test asks about a parabola. The numbers in the working are found by the engine.

Example.

1. Reading the vertex form. The function $f$ is defined by $f(x) = -2(x + 3)^2 + 8$ . What is the vertex of its graph, and is it a maximum or a minimum?

Match the function to $a(x - h)^{2} + k$ . The parentheses hold $x + 3$ , which is $x - (-3)$ , so $h$ is $-3$ . The number outside is $k$ , 8. Put $-3$ in to confirm, and try one point on each side:

\begin{aligned} f(x) &= -2(x + 3)^{2} + 8 \\ k &= f(h) \\ &= -2 \times (-3 + 3)^{2} + 8 \\ &= 8 \\ L &= f(-4) \\ &= -2 \times (-4 + 3)^{2} + 8 \\ &= 6 \\ R &= f(-2) \\ &= -2 \times (-2 + 3)^{2} + 8 \\ &= 6 \end{aligned}

The vertex is $({-3}, {8})$ . The value on each side is $6$ , which is lower, and $a$ is $-2$ , negative: the parabola opens down and $8$ is the maximum.

Example.

2. From the standard form. For what value of $x$ does $y = x^2 - 10x + 7$ reach its minimum, and what is the minimum?

Here $a$ is 1 and $b$ is $-10$ , so $-b$ is 10. The $x$ of the vertex is $-b$ divided by $2a$ . Then put that $x$ into the function.

\begin{aligned} h &= \frac{10}{2a} \\ &= \frac{10}{2 \times 1} \\ &= 5 \\ k &= h^{2} - 10h + 7 \\ &= 5^{2} - 10 \times 5 + 7 \\ &= -18 \end{aligned}

The minimum is $-18$ , and it is reached at $x$ $=$ $5$ . If the question asks "for what value of $x$ ", the answer is $5$ , not $-18$ . The vertex form confirms both numbers at once:

\begin{aligned} y &= (x - 5)^{2} - 18 \\ &= x^{2} - 10x + 7 \end{aligned}

Example.

3. From the factored form. The function $f$ is defined by $f(x) = (x + 2)(x - 8)$ . For what value of $x$ is $f(x)$ least?

The factored form shows the zeros: $f$ is 0 at $x = -2$ and at $x = 8$ . The parabola is symmetric, so the vertex is halfway between them.

\begin{aligned} h &= \frac{r + s}{2} \\ &= \frac{-2 + 8}{2} \\ &= 3 \\ f(x) &= (x + 2)(x - 8) \\ k &= f(h) \\ &= (3 + 2) \times (3 - 8) \\ &= -25 \end{aligned}

$f(x)$ is least at $x$ $=$ $3$ , where its value is $-25$ . Neither zero is the answer, and neither is $-16$ , the product of $-2$ and 8: that is the value at $x = 0$ .

Example.

4. The vertex in a story. A bakery sells a loaf for $p$ dollars. Its daily profit from the loaves, in dollars, is modeled by $P = -20(p - 6)^2 + 320$ . What does the vertex of the graph mean, and at what prices is the profit zero?

The model is in vertex form with $h$ 6 and $k$ 320, and $a$ is negative, so the vertex $(6, 320)$ is a maximum: a price of 6 dollars gives the greatest daily profit, 320 dollars. The profit is zero where $P$ is 0:

\begin{aligned} -20(p - 6)^{2} + 320 &= 0 \\ p &= [2, 10] \\ r &= \operatorname{min}(p) \\ &= 2 \\ s &= \operatorname{max}(p) \\ &= 10 \\ h &= \frac{r + s}{2} \\ &= \frac{2 + 10}{2} \\ &= 6 \end{aligned}

The profit is zero at a price of $2$ dollars and at $10$ dollars, and negative outside them. The best price, $6$ , is halfway between the two, as the symmetry requires.

Example.

5. A shifted function. The function $f$ is defined by $f(x) = x^2 - 6x + 1$ , and $g(x) = f(x + 4)$ . For what value of $x$ does $g(x)$ reach its minimum?

First find where $f$ is least: in $f$ , $b$ is $-6$ , so $-b$ is 6. Then ask which $x$ makes the input of $f$ equal to that number, because $g$ hands $x + 4$ to $f$ .

\begin{aligned} h &= \frac{6}{2 \times 1} \\ &= 3 \\ x + 4 &= h \\ x + 4 &= 3 \\ m &= -1 \end{aligned}

$g$ is least at $x$ $=$ $-1$ . Adding 4 inside the function moves the graph 4 to the left, not to the right. Writing $g$ out shows the same vertex:

\begin{aligned} y &= (x + 4)^{2} - 6(x + 4) + 1 \\ &= x^{2} + 2x - 7 \\ m &= \frac{-2}{2 \times 1} \\ &= -1 \end{aligned}

Example.

6. A model told in words. A ball is kicked from the ground. It reaches its greatest height, 36 feet, after 1.5 seconds, and its height is a quadratic function of time. How high is the ball 2 seconds after the kick?

The greatest height is the vertex, so write the vertex form $a(t - 1.5)^{2} + 36$ with $a$ still unknown. The ball starts on the ground: the height is 0 when $t$ is 0. That gives $a$ .

\begin{aligned} a(0 - h)^{2} + k &= 0 \\ a(0 - \frac{3}{2})^{2} + 36 &= 0 \\ a &= -16 \\ f(t) &= a(t - h)^{2} + k \\ &= -16(t - \frac{3}{2})^{2} + 36 \\ y &= f(2) \\ &= -16 \times (2 - \frac{3}{2})^{2} + 36 \\ &= 32 \end{aligned}

After 2 seconds the ball is $32$ feet up. By symmetry it is at the same height at 1 second, and it lands at 3 seconds, as far after the vertex as the kick was before it.

\begin{aligned} y &= f(1) \\ &= -16 \times (1 - \frac{3}{2})^{2} + 36 \\ &= 32 \\ y &= f(3) \\ &= -16 \times (3 - \frac{3}{2})^{2} + 36 \\ &= 0 \end{aligned}

Practice

Answer SAT-style questions on vertex form and graphing parabolas. Each answer is checked, with the full solution, and every try brings new numbers. Practice needs a free account: it keeps your progress.

Practise vertex form and graphing parabolas with new numbers