Question

How do you graph by using the zeros for `f(x)=-4x^3+4x^2+15x`?

Answer 1

`x=0,x=-3/2,x=5/2`

Explanation:

To find the zeroes, we'll first factor out an `x`:
`x(-4x^2+4x+15)`

We can then factor the quadratic by grouping:
`x(-4x^2-6x+10x+15)`

`x(-2x(2x+3)+5(2x+3))`

`x(-2x+5)(2x+3)`

Now we can use the zero factor principle to figure out that the zeroes are:
`x=0,x=-3/2,x=5/2`

In terms of the graph we can say that the graph will go from positive infinity (because of the negative leading coefficient), cross the `x`-axis and go negative at `x=-3/2` (and it will cross, since the multiplicity is odd) then cross the `x`-axis again to become positive at `x=0`, and finally go negative again at `x=5/2` to continue on to negative infinity.

We can see that our conclusions were in fact correct if we look at the graph:
graph{-4x^3+4x^2+15x [-35.9, 29.07, -9.97, 22.5]}