Question

find the max of: f(x)=x-2x^2-1

Answer 1

its a quadratic function btw

Answer 2

Know how to find the places where the slope is 0?

Answer 3

yeah but i usually find it using a data table

Answer 4

is there a faster way of doing that?

Answer 5

A table works too. Sometimes you can find the x intercepts and see what the y for half way between them is. In this case, it has no intercepts. As it is a quadratic you can find the vertex.

Answer 6

yeah ik

Answer 7

im just having some trouble finding the max (highest vertex) of this function

Answer 8

If you can get it into the form \(y = a(x – h)^2 + k\), then (h, k) is the vertex. But if you can not see any easy path to that form, here is a page that shows the formula for finding it in any case:

http://www.purplemath.com/modules/sqrvertx2.htm

If you look there, the x point is \(-\dfrac{b}{2a}\). Once you have that, you can plug it in to get the y value.

Answer 9

Whee... let me try that formula again.

\(y=a(x+h)^2-k\) got messed up.

Answer 10

notice the leading term is a negative one... so that means the "parabolic" graph is opening "downwards"

thus the vertex is at the peek of it \(\bf f(x)=x-2x^2-1\implies f(x)={\color{red}{ -2}}x^2{\color{blue}{ +1}}x{\color{green}{ -1}}
\\ \quad \\

\textit{vertex of a parabola}\\ \quad \\



\left(-\cfrac{{\color{blue}{ b}}}{2{\color{red}{ a}}}\quad ,\quad {\color{green}{ c}}-\cfrac{{\color{blue}{ b}}^2}{4{\color{red}{ a}}}\right)\)

Answer 11

Also, remember you need it in standard form:

\(ax^2+bx+c\)

Yours is out of order, so a is not where it should be.

Answer 12

peak rather =)

Answer 13

Yep

Answer 14

^ - ^ thanks lol