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

1 Answer

3 votes

Answer:


y=(x-3)^(2) -25

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=(-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=(-b)/(2a)
h=(-(-6))/(2(1))=(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!

User Aaron Daniels
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