Question

How do you find the intercepts, vertex and graph `f(x)=-0.25x^2-3x`?

Answer 1

Vertex is `(6,-9)`
Intercepts are `(0,0)` and `(12,0)`

Explanation:

`f(x)=0.25x^2-3x`
Calculate first derivative `f'(x)=0.5x-3`
Let `f'(x)=0`
So `(0.5x-3=0) =>x=3/0.5=6`
Calculate the second derivative `f''(x)=0.5 >0`
So we have a minimum which is `(6,-9)`
Intercept with the y- axis when `x=0=>y=0` So we have the point `(0,0)`
Intercept with the y-axis when `y=0`
`0.25x^2-3x=0 =>x(0.25x-3)=0`
So `x=0`and `x=3/0.25=12`
So the point is `(12,0)`
Here is a sketch of the graph

graph{0.25x^2-3*x [-20, 20, -10, 10]} /