F(x)= x^2+8x

Express the quadratic function in standard form

Question

F(x)= x^2+8x

Express the quadratic function in standard form

zepdrix · Answer

Hmm so we have a parabola, I think standard form looks something like this if I remember correctly:
$$f(x)=(x-h)^2 + k$$
So we need to complete the square, then we'll have a left over quantity (k).
Make sense? :D

anonymous · Answer

whats my x and whats my h?

zepdrix · Answer

It'll make a little more sense once you see the answer :)
First step is to complete the square,
So we take half of the b term (half of 8) and square it.

anonymous · Answer

x^2+8x+16

zepdrix · Answer

$$f(x)=x^2+8x+16-16$$
Ok good! :) but we can't just add 16 willy nilly! we have to balance the equation like this.

anonymous · Answer

thats the same as x^2+8x

zepdrix · Answer

Yes, but we wont let the 16's cancel out, we're going to do something with them ^^

zepdrix · Answer

x^2+8x+16 is a perfect square, what does it simply into?

anonymous · Answer

(x+4)^2

zepdrix · Answer

$$f(x)=(x^2+8x+16)-16$$$$f(x)=(x+4)^2-16$$
Ok very good :) and we still have the 16 left over right?
Is it a litte easier to recognize your H and K now?

zepdrix · Answer

Uh ohhhh, Brent confused? :D

anonymous · Answer

h=-4   
k=-16

zepdrix · Answer

Ok good good :D
And what is significant about that point is the fact that it tells us where our vertex is located!
That center point from which the parabola grows in 2 directions :D

anonymous · Answer

i need to put it in standard form and find the vertex and x/y intercepts

zepdrix · Answer

$$f(x)=(x+4)^2-16$$
Ok so we've gotten it into standard form.
You identified h and k.
(h,k) is your vertex.

Now to find intercepts, do the following:

For y-intercepts, plug in x=0 and solve for f(x).

For x-intercepts, plug in f(x)=o and solve for x (you should get 2 values for this one).

anonymous · Answer

so then there is a y-int at 0

zepdrix · Answer

the y-axis is intercepted (crossed by our function) when x=0.
When we plug x=0 into our function, we get:
$$f(x)=(0+4)^2-16$$$$f(x)=0$$
Ah so we do! :) Yes good job.

anonymous · Answer

what about the x?

zepdrix · Answer

We'll plug 0 in for the other variable and solve for x.

anonymous · Answer

X^2+2*x*4+4^2-4^2, so (x+4)^2-16

zepdrix · Answer

$$0=(x+4)^2-16$$
Solve for x from here. Important note, when you take the square root of both sides, you'll actually get 2 values for (x+4)! :D

Ok i gotta go, test tomorrow :D gotta get some sleep.
Good luck with the rest of the problem Brent!

anonymous · Answer

its 0!

anonymous · Answer

so x has an intercept at zero too