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

Jan 1, 2018

$x = 0 , x = - \frac{3}{2} , x = \frac{5}{2}$

#### Explanation:

To find the zeroes, we'll first factor out an $x$:
$x \left(- 4 {x}^{2} + 4 x + 15\right)$

We can then factor the quadratic by grouping:
$x \left(- 4 {x}^{2} - 6 x + 10 x + 15\right)$

$x \left(- 2 x \left(2 x + 3\right) + 5 \left(2 x + 3\right)\right)$

$x \left(- 2 x + 5\right) \left(2 x + 3\right)$

Now we can use the zero factor principle to figure out that the zeroes are:
$x = 0 , x = - \frac{3}{2} , x = \frac{5}{2}$

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 $x = - \frac{3}{2}$ (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 $x = \frac{5}{2}$ 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]}