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

1 Answer
Jan 1, 2018

Answer:

#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]}