Question

at what points are the tangents to the graph of f(x)=x^2-3x horizontal?

Answer 1

set the derivative equal to zero and solve

Answer 2

\[f(x)=2x^2-3x\]
\[f'(x)=2x-3\] set it equal to zero and solve
you don't really need calc for this.
this is a quadratic, and slope is zero at the vertex
first coordinate of the vertex is \(-\frac{b}{2a}\) which is what you get when you take the derivative, set equal zero and solve

Answer 3

ok so i get (-3/2)... im also supposed to somehow get the point (-9/4). how do i do that?

Answer 4

you have the function right?

Answer 5

it is \((\frac{3}{2},f(\frac{3}{2}))\)

Answer 6

oh btw it is
\[2x-3=0\implies x=\frac{3}{2}\] not negative

Answer 7

oh sorry thats what i meant! and i get it! thanksss!

Answer 8

yw