Question

Consider the quadratic function y=x^2-8x+195. How do I write the function in vertex form? How do I identify the vertex?

Answer 1

Completing the square method would work here.

Answer 2

I have to complete the square also, but I still have to write it in vertex form and I don't know how to

Answer 3

You start by completing the square

Answer 4

how do I do that

Answer 5

Vertex form of the quadratic is: \(y=a(x-h)^2 +k\)

Answer 6

So we start by grouping, put a parenthesis around the x-terms

Answer 7

h = -b/(2a)
k = f(-b/(2a))

Answer 8

and then plug in (0, 195) to get 'a'

Answer 9

What is f?

Answer 10

\[y=(x^2-8x)+195\]\[c=\left(\frac{b}{2}\right)^2 = \left(\frac{-8}{2}\right)^2 = (-4)^2 = 16\]\[y=(x^2-8x+16)+195-16\]\[y=(x-4)^2+179\]

Answer 11

Basically, when you are completing the square, you are "completing" a quadratic inside the parenthesis. You find a `new` c value that is added to both the quadratic and the constant outside the quadratic, respectively.

Answer 12

how do I find out what the vertex is?

Answer 13

\[y=a(x-\color{blue}h)^2 +\color{blue}k\]\[y=1(x-\color{blue}4)^2+\color{blue}{179}\]\[\text{vertex of parabola} = (h,k)\]

Answer 14

Make sense?

Answer 15

so it would be (4, 179)? Not -4?

Answer 16

how would I know if it has any x intercepts?

Answer 17

Yes, (4,179)

Answer 18

Ok, I do understand how to do that now...how would I know if it has any x intercepts?

Answer 19

\[(x-4)^2+179=0\] Solve for x

Answer 20

Thank you very much

Answer 21

No problem :)