Transform the function in vertex form

1. f(x)= x^2 + 8x - 6

Question

Transform the function in vertex form

1. f(x)= x^2 + 8x - 6

precal · Answer

you will need to complete the square

ash2326 · Answer

A parabola with vertex (h, k) has the equation
$$y=a(x-h)^2+k$$
We need to transform y=x^2-8x-6 in the above form. Can you try now?

anonymous · Answer

what does h and k stand for?

ash2326 · Answer

(h, k) are the coordinates of the vertex of a parabola|dw:1336921559619:dw|

anonymous · Answer

ok

ash2326 · Answer

Can you try now?

anonymous · Answer

i still dont get it

ash2326 · Answer

we have 
$$f(x)=x^2+8x-6$$
and 
we need to convert it to 
$$f(x)=a(x-h)^2+k$$
We need to find a, x and h, Let's try to find a
We notice that in the equation $f(x)=x^2+8x-6$, coefficient of x is 1 so
$$a=1$$
Do you get this?

ash2326 · Answer

* coefficient of x^2

anonymous · Answer

You have to "complete the square" of f(x)=x^2+8x-6
$$f(x)=(x+4)^2-22$$So your vertex is at the point (-4,-22).

anonymous · Answer

Let me know if you are unsure how to do this

anonymous · Answer

Basically it works like this:$$f(x)=ax^2+bx+c$$transforms to$$f(x)=a(x-h)^2+k$$The coefficent on x^2 (=a) stays the same in both equations.  To get the "h" in the second equation we take h=b/2.  Now all that's left is the "k" value.  Our k is found by noting that h^2+k=c...so k=c-h^2.

precal · Answer

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