Question

find the minimum y-value on the graph of y=f(x)
f(x)=-x^2+4x+8

Answer 1

Have you taken calculus yet or are you in pre-calc/algebra?

Answer 2

algebra yes no cal

Answer 3

Okay. Do you know how to complete the square? Doing that, you can rewrite the parabola in vertex form.

Answer 4

i understand some of it but i usually screw some thing up in it this is very hard for me i have never been any good at higher math

Answer 5

\[\begin{align*}f(x)&=-x^2+4x+8\\
&=-\left(x^2-4x-8\right)\\
&=-\left(x^2-4x+4-12\right)\\
&=-\left((x-2)^2-12\right)\\
&=(x-2)^2+12\end{align*}\]
So now f(x) is in vertex form,
\[a(x-h)^2+k, \text{ where $(h,k)$ is the vertex.}\]

Answer 6

ok i understand that part of it

Answer 7

Oh sorry, that should be
\[-(x-2)^2+12\]

Answer 8

The thing is, now that the value of a is negative, the parabola opens downward, so the vertex represents the maximum value of the function.

Answer 9

ok I understand the lanquage of it it is the doing of it i dont understand

Answer 10

You mean how to come up with the numbers in order to rewrite f(x)?

Answer 11

yes how to work the problem i get confused with steps

Answer 12

Okay. Is it alright if I try to explain the process using variables (a, b, and c) or would you prefer something more concrete (actual number coefficients)?

Answer 13

a b and c work for me

Answer 14

Alright
\[\begin{align*}y&=ax^2+bx+c\\
y&=a\left(x^2+\frac{b}{a}x+\frac{c}{a}\right)\\
y&=a\left(x^2+\frac{b}{a}x+\left(\frac{b^2}{(2a)^2}-\frac{b^2}{(2a)^2}\right)+\frac{c}{a}\right)
\end{align*}\]
Do you follow so far?