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]}

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