Question

Y=x^2-6x-16 in vertex form

Answer 1

Answer:

$y=(x-3)^{2} -25$

Step-by-step explanation:

The standard form of a quadratic equation is $y=ax^{2} +bx+c$

The vertex form of a quadratic equation is $y=a(x-h)^{2} +k$

The vertex of a quadratic is (h,k) which is the maximum or minimum of a quadratic equation. To find the vertex of a quadratic, you can either graph the function and find the vertex, or you can find it algebraically.

To find the h-value of the vertex, you use the following equation:

$h=\frac{-b}{2a}$

In this case, our quadratic equation is $y=x^{2} -6x-16$. Our a-value is 1, our b-value is -6, and our c-value is -16. We will only be using the a and b values. To find the h-value, we will plug in these values into the equation shown below.

$h=\frac{-b}{2a}$ ⇒ $h=\frac{-(-6)}{2(1)}=\frac{6}{2} =3$

Now, that we found our h-value, we need to find our k-value. To find the k-value, you plug in the h-value we found into the given quadratic equation which in this case is $y=x^{2} -6x-16$

$y=x^{2} -6x-16$ ⇒ $y=(3)^{2} -6(3)-16$ ⇒ $y=9-18-16$ ⇒ $y=-25$

This y-value that we just found is our k-value.

Next, we are going to set up our equation in vertex form. As a reminder, vertex form is: $y=a(x-h)^{2} +k$

a: 1

h: 3

k: -25

$y=(x-3)^{2} -25$

Hope this helps!