How do you graph by using the zeros for g(x)=1/10(x+1)^2(x-3)^3?

Aug 12, 2018

x intercepts : $x = - 1 , x = 3$, y intercept: $y = - 2.7$
local minimum at $\left(0.6 , - 3.54\right)$

Explanation:

$g \left(x\right) = \frac{1}{10} {\left(x + 1\right)}^{2} {\left(x - 3\right)}^{3}$

Vertical asymptote occur when denominator is zero.

So no vertical assymptote. Degree is ${x}^{5}$ , odd and

leading coefficient is $\left(+\right)$ , so end behavior is "down /up"

i.e $y \to - \infty$ as $x \to - \infty$ and

$y \to \infty$ as $x \to \infty$

Zeros are $x = - 1$ with multiplicity of $2$ and

$x = 3$ with multiplicity of $3 \therefore$ x intercepts are

$x = - 1 \mathmr{and} x = 3$, y intercept: Putting $x = 0$ in the equation

we get,$y = \frac{1}{10} \cdot {\left(1\right)}^{2} \cdot {\left(- 3\right)}^{3} = - 2.7$

$g \left(x\right) = \frac{1}{10} {\left(x + 1\right)}^{2} {\left(x - 3\right)}^{3}$

${g}^{'} \left(x\right) = \frac{1}{10} \left[2 \left(x + 1\right) {\left(x - 3\right)}^{3} + {\left(x + 1\right)}^{2} \cdot 3 {\left(x - 3\right)}^{2}\right]$ or

g^'(x)= 1/10[(x+1)(x-3)^2(2(x-3)+3(x+1)] or

${g}^{'} \left(x\right) = \frac{1}{10} \left[\left(x + 1\right) {\left(x - 3\right)}^{2} \left(5 x - 3\right)\right]$

$g ' \left(x\right) = 0 \therefore$ slope is zero at $x = - 1 , x = 3 \mathmr{and} x = \frac{3}{5} = 0.6$

$g \left(0.6\right) \approx - 3.5389 \approx - 3.54$ , critical point $x = 0.6$

Slope at $x < - 0.6 , {g}^{'} \left(x\right) < 0$ i.e decreasing and

slope at $x > - 0.6 , {g}^{'} \left(x\right) > 0$ i.e increasing.

So $\left(0.6 , - 3.54\right)$ is local minimum.

At $x = - 1$ the graph will not cross $x$ axis as the multiplicity

is even $2$ and at $x = 3$ the graph will cross $x$axis as the

multiplicity is odd $2$.

graph{1/10(x+1)^2(x-3)^3 [-10, 10, -5, 5]}[Ans]