How do you use end behavior, zeros, y intercepts to sketch the graph of #f(x)=-7x^3-5x^2+4x#?

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Answer 1 · 2017-12-05T21:09:04

See below.

y axis intercepts when #x=0#:

#y=-7(0)^3-5(0)^2+4(0)=0#

Coordinate: # ( 0 , 0 )#

x axis intercept when #y=0#:

#-7x^3-5x^2+4x=0#

Factor:

#x(-7x^2-5x+4)=0=>x=0#

#7x^2+5x-4=0#

Using quadratic formula:

#x=(-5+-sqrt(137))/(14)=(-5+sqrt(137))/14 and (-5-sqrt(137))/14#

x intercepts at:

#( 0 , 0 ) , ( (-5+sqrt(137))/14 , 0 ) and ((-5-sqrt(137))/14, 0 )#

For end behaviour of polynomials we only need to look at the degree and leading coefficient:

as #x->oo# , #color(white)(88)-7x^3->-oo#

as #x->-oo# , #color(white)(88)-7x^3->oo#

Graph:

graph{y=-7x^3-5x^2+4x [-10, 10, -5, 5]}