Lesson Objectives
• Learn how to find the vertex form of a parabola

## How to Find the Vertex Form of a Parabola

In this lesson, we will go deeper into the topic of quadratic functions and learn how to find the vertex form. Note: In some textbooks or other resources, the "vertex form" is referred to as the "standard form".

A quadratic function is of the form: $$f(x) = ax^2 + bx + c$$ where a, b, and c are real numbers, with a ≠ 0
The graph of any quadratic function is known as a parabola and can be obtained from the graph of f(x) = x2 using the function transformations that we previously studied. In the next lesson, we will learn how to sketch the graph of a parabola. For now, we will only focus on finding the vertex form of the parabola, which is useful for graphing and other scenarios.

### Vertex Form of a Quadratic Function

$$f(x) = a(x - h)^2 + k$$ When our quadratic function is in vertex form, we can state the vertex occurs at (h, k). The vertex of a parabola is the lowest point when the parabola opens up and the highest point when the parabola opens down.
The parabola opens up if a > 0: For the graph above, the vertex occurs at (4, 1). This is the lowest point on the graph since the parabola opens up.
The parabola opens down if a < 0: For the graph above, the vertex occurs at (4, 1). This is the highest point on the graph since the parabola opens down.
We can obtain the vertex form by completing the square or using the vertex formula. $$f(x) = ax^2 + bx + c$$ Group the variable terms together using parentheses: $$f(x) = (ax^2 + bx) + c$$ Factor out a, the coefficient of x2: $$f(x) = a\left(x^2 + \frac{b}{a}x\right) + c$$ Multiply the coefficient of x, b/a by 1/2, and then square the result: $$\left(\frac{b}{a} \cdot \frac{1}{2}\right)^2 = \left(\frac{b}{2a}\right)^2 = \frac{b^2}{4a^2}$$ Add and subtract this result on the inside of the parentheses: $$f(x) = a\left(x^2 + \frac{b}{a}x + \frac{b^2}{4a^2} - \frac{b^2}{4a^2}\right) + c$$ Distribute the a to the final term inside of the parentheses and close down the parentheses: $$f(x) = a\left(x^2 + \frac{b}{a}x + \frac{b^2}{4a^2}\right) - \frac{ab^2}{4a^2} + c$$ Factor the perfect square trinomial, simplify, and rearrange: $$f(x) = a\left(x + \frac{b}{2a}\right)^2 + c - \frac{b^2}{4a}$$ Match the vertex form: $$f(x) = a(x - h)^2 + k$$ $$f(x) = a\left(x - \left(-\frac{b}{2a}\right)\right)^2 + c - \frac{b^2}{4a}$$ $$h = -\frac{b}{2a}$$ $$k = c - \frac{b^2}{4a}$$ We can also write k as: $$k = f(h)$$

### Vertex Formula

The result above can be used to find the vertex form of any quadratic equation. $$f(x) = ax^2 + bx + c, a ≠ 0$$ Vertex Form: $$f(x) = a(x - h)^2 + k, a ≠ 0$$ $$h = -\frac{b}{2a}$$ $$k = f(h)$$ Let's look at an example.
Example #1: Find the vertex form. $$f(x) = 4x^2 + 56x + 197$$ We can do this by completing the square or by using the vertex formula. Let's start with the completing the square method. $$f(x) = 4x^2 + 56x + 197$$ Group the variable terms together using parentheses: $$f(x) = (4x^2 + 56x) + 197$$ Factor out 4, the coefficient of x2: $$f(x) = 4(x^2 + 14x) + 197$$ Multiply the coefficient of x, 14 by 1/2, and then square the result: $$\left(14 \cdot \frac{1}{2}\right)^2 = 7^2 = 49$$ Add and subtract this result on the inside of the parentheses: $$f(x) = 4(x^2 + 14x + 49 - 49) + 197$$ Distribute the 4 to the final term inside of the parentheses and close down the parentheses: $$f(x) = 4(x^2 + 14x + 49) - 49 \cdot 4 + 197$$ Factor the perfect square trinomial and simplify: $$f(x) = 4(x + 7)^2 + 1$$ If you were asked for the vertex (h, k), then you can match the vertex form: $$f(x) = a(x - h)^2 + k$$ $$f(x) = 4(x - (-7))^2 + 1$$ $$h = -7$$ $$k = 1$$ $$\text{Vertex:}\, (-7, 1)$$ Alternatively, we could just use the vertex formula given above.
Vertex Formula: $$h=-\frac{b}{2a}$$ $$k=f\left(h\right)$$ Recall that a is the coefficient of the squared variable and b is the coefficient of the variable raised to the first power.
$$a=4$$ $$b=56$$ To find h, let's plug into the formula: $$h=-\frac{b}{2a}$$ $$h=-\frac{56}{8}$$ $$h=-7$$ To find k, we find f(-7): $$f(-7) = 4(-7)^2 + 56(-7) + 197$$ $$=4(49) + 56(-7) + 197$$ $$=196 - 392 + 197$$ $$=-196 + 197 = 1$$ Now, we fill in the vertex form: $$f(x)=a(x - h)^2 + k$$ $$a=4, h=-7, k=1$$ $$f(x) = 4(x - (-7))^2 + 1$$ $$f(x)=4(x + 7)^2 + 1$$ There is really no need to complete the square each time, using the vertex formula once you understand where it comes from is the normal way to solve this type of problem. Let's look at a few more examples using only the vertex formula.
Example #2: Find the vertex form. $$f(x) = -3x^2 - 6x - 11$$ $$a = -3$$ $$b = -6$$ $$h = -\frac{b}{2a} = -\frac{-6}{2(-3)}= -\frac{-6}{-6} = -1$$ $$k = f(-1)$$ $$f(-1) = -3(-1)^2 - 6(-1) - 11$$ $$=-3 + 6 - 11 = 3 - 11 = -8$$ $$k = -8$$ $$\text{Vertex:}\, (-1, -8)$$ Vertex Form: $$f(x)=a(x - h)^2 + k$$ $$a = -3, h = -1, k = -8$$ $$f(x) = -3(x - (-1))^2 + (-8)$$ $$f(x) = -3(x + 1)^2 - 8$$ Example #3: Find the vertex form. $$f(x) = -\frac{1}{4}x^2 - \frac{3}{2}x - \frac{13}{4}$$ $$a = -\frac{1}{4}$$ $$b = -\frac{3}{2}$$ $$h = -\frac{b}{2a} = -\left(\frac{-\frac{3}{2}}{2 \cdot -\frac{1}{4}}\right) = -\left(-\frac{3}{2} \cdot -\frac{2}{1}\right) = -3$$ $$k = f(-3)$$ $$f(-3) = -\frac{1}{4}(-3)^2 - \frac{3}{2}(-3) - \frac{13}{4}$$ $$= -\frac{9}{4} + \frac{9}{2} - \frac{13}{4}$$ $$= -\frac{9}{4} + \frac{18}{4} - \frac{13}{4}$$ $$= \frac{-9 - 13 + 18}{4} = -1$$ $$k = -1$$ $$\text{Vertex:}\, (-3, -1)$$ Vertex Form: $$f(x)=a(x - h)^2 + k$$ $$a = -\frac{1}{4}, h = -3, k = -1$$ $$f(x) = -\frac{1}{4}(x - (-3))^2 + (-1)$$ $$f(x) = -\frac{1}{4}(x +3)^2 - 1$$

#### Skills Check:

Example #1

Write in vertex form. $$f(x)=-2x^2 + 36x - 157$$

A
$$f(x)=-2(x - 9)^2 - 5$$
B
$$f(x)=-2(x - 9)^2 + 5$$
C
$$f(x)=2(x - 8)^2 + 3$$
D
$$f(x)=(x - 7)^2 + 9$$
E
$$f(x)=2(x - 4)^2 + 1$$

Example #2

Write in vertex form. $$f(x)=-7x^2 - 56x - 110$$

A
$$f(x)=-7(x + 4)^2 + 2$$
B
$$f(x)=(x + 5)^2 + 1$$
C
$$f(x)=7(x + 4)^2 + 2$$
D
$$f(x)=-7(x + 1)^2 + 6$$
E
$$f(x)=(x - 1)^2 + 3$$

Example #3

Write in vertex form. $$f(x)=-\frac{1}{2}x^2 + 7x - \frac{47}{2}$$

A
$$f(x)=-\frac{1}{2}(x - 7)^2 + 1$$
B
$$f(x)=-\frac{1}{2}(x + 9)^2 + 1$$
C
$$f(x)=-\frac{1}{2}(x + 3)^2 - 1$$
D
$$f(x)=-2(x - 3)^2 - 1$$
E
$$f(x)=-2(x + 7)^2 - 3$$