Find the vertex and the intercept of the parabola: f(x) = x2 - x - 2?

Answer 1

Rachel, you need to learn to show exponents! This is the 3rd problem I've seen of yours where you left out the ^!

Your equation is f(x) = x^2 - x - 2

The vertex is the point where the tangent is zero, so take the derivative and set it to zero:

f'(x) = 2x-1, so x = 1/2

Plug into the original equation to get the f(1/2) =
(1/2)^2 - 1/2 - 2 = -9/4

So your vertex is at the point (1/2,-9/4)

Answer 2

f(x)=(x-.5)^2-2.25
Therefore, the vertex occurs at (.5,-2.25)

Answer 3

do your homework yourself...