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

Question

anonymous · Answer

set the derivative equal to zero and solve

anonymous · Answer

$$f(x)=2x^2-3x$$
$$f'(x)=2x-3$$ 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

anonymous · Answer

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

anonymous · Answer

you have the function right?

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

yw