identify the vertex and intercepts of f(x)=1/4(x^2-16x+32)

[unregistered]

I was using this awesome formula for completing the square that the someone gave me in this forum and it was working flawlessly until I got to this equation.

\(\displaystyle f(x)=\frac{1}{4}(x^{2}-16x+32)\) (original equation)

\(\displaystyle f(x)=\frac{1}{4}(x+\frac{16}{2(1/4)})^{2}+(32-\frac{16^{2}}{4(1/4})\)

\(\displaystyle f(x)=\frac{1}{4}(x+32)^{2}+(32-\frac{256}{1})\)

\(\displaystyle f(x)=\frac{1}{4}(x+32)^{2}+(-224)\)

My graphing calculator shows the vertex at (8,-8) so I'm not sure if there is anything further I need to do or if I just did something wrong. I have to get the vertex before I can get the intercept so I'm stuck here.

Thanks for any help.

 

[royhaas]

\(\displaystyle f(x) = \frac{1}{4}(x^2-16x+32) =\frac{1}{4}(x^2-16x+64-32)=\frac{1}{4}\{(x-8)^2 -32\}\).

 

[unregistered]

So I cannot use the completing the square formula for this? Is there a rule that I can follow for when to and when not to use the formula?

 

[royhaas]

You did not use the formula correctly.

 

[unregistered]

I have memorized this formula, it is still:

\(\displaystyle a(x-\frac{b}{2a})^2+(c-\frac{b^2}{4a})\)

yes?

I substituted my values for the ones in this formula so what did I do wrong? Sorry to keep pressing, I just really want to understand.

 

[stapel]

I have memorized this formula, it is still:

\(\displaystyle a(x-\frac{b}{2a})^2+(c-\frac{b^2}{4a})\)

To apply this formula, you must start with the "ax² + bx + c" form, which you did not.

To convert what you had to the required starting form, multiply the "1/4" through the parentheses. Then, once you know the values for "a", "b", and "c", you can then apply the formula.

Eliz.

 

[unregistered]

AWESOME, that is what I was looking for, I had to factor out the leading coefficient and put into standard form, that is what I totally needed, so now I can finish with the formula that Glactus gave me.

\(\displaystyle f(x)=\frac{1}{4}(x^{2}-16x+32)\) (original equation)

\(\displaystyle f(x)=1/4x^{2}-4x+8\) (factor out the leading coefficient \(\displaystyle 1/4\))

\(\displaystyle f(x)=1/4(x+\frac{-4}{2(1/4)})^{2}+(8-\frac{-4^{2}}{4(1/4})\)

\(\displaystyle f(x)=1/4(x+\frac{-4}{1/2})^{2}+(8-\frac{-4^{2}}{1})\)

\(\displaystyle f(x)=1/4(x+(-8))^{2}+(8-16)\)

\(\displaystyle f(x)=1/4(x-8)^{2}+(-8)\)

vertex (8,-8)

Thanks Eliz & Royhaas.