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

Dec 5, 2017

See below.

#### Explanation:

y axis intercepts when $x = 0$:

$y = - 7 {\left(0\right)}^{3} - 5 {\left(0\right)}^{2} + 4 \left(0\right) = 0$

Coordinate: $\left(0 , 0\right)$

x axis intercept when $y = 0$:

$- 7 {x}^{3} - 5 {x}^{2} + 4 x = 0$

Factor:

$x \left(- 7 {x}^{2} - 5 x + 4\right) = 0 \implies x = 0$

$7 {x}^{2} + 5 x - 4 = 0$

$x = \frac{- 5 \pm \sqrt{137}}{14} = \frac{- 5 + \sqrt{137}}{14} \mathmr{and} \frac{- 5 - \sqrt{137}}{14}$

x intercepts at:

$\left(0 , 0\right) , \left(\frac{- 5 + \sqrt{137}}{14} , 0\right) \mathmr{and} \left(\frac{- 5 - \sqrt{137}}{14} , 0\right)$

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

as $x \to \infty$ , $\textcolor{w h i t e}{88} - 7 {x}^{3} \to - \infty$

as $x \to - \infty$ , $\textcolor{w h i t e}{88} - 7 {x}^{3} \to \infty$

Graph:

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