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Marina86 [1]
3 years ago
11

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

Mathematics
1 answer:
satela [25.4K]3 years ago
3 0

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!

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