STEP 1:Find the x-intercepts

To find the x-intercepts solve, the equation $ \color{blue}{ -10x^5+10x^4+18x^3 = 0 } $

The solutions of this equation are:

$$ \begin{matrix}x_1 = 0 & x_2 = 1.9318 & x_3 = -0.9318 \end{matrix} $$

(you can use the step-by-step polynomial equation solver to see a detailed explanation of how to solve the equation)

STEP 2:Find the y-intercepts

To find the y-intercepts, substitute $ x = 0 $ into $ \color{blue}{ p(x) = -10x^5+10x^4+18x^3 } $, so:

$$ \text{Y inercept} = p(0) = 0 $$

STEP 3:Find the end behavior

The end behavior of a polynomial is the same as the end behavior of a leading term.

$$ \lim_{x \to -\infty} \left( -10x^5+10x^4+18x^3 \right) = \lim_{x \to -\infty} -10x^5 = \color{blue}{ \infty } $$

The graph starts in the upper-left corner.

$$ \lim_{x \to \infty} \left( -10x^5+10x^4+18x^3 \right) = \lim_{x \to \infty} -10x^5 = \color{blue}{ -\infty } $$

The graph ends in the lower-right corner.

STEP 4:Find the turning points

To determine the turning points, we need to find the first derivative of $ p(x) $:

$$ p^{\prime} (x) = -50x^4+40x^3+54x^2 $$

The x coordinate of the turning points are located at the zeros of the first derivative

$$ p^{\prime} (x) = 0 $$ $$ \begin{matrix}x_1 = 0 & x_2 = 1.5136 & x_3 = -0.7136 \end{matrix} $$

(cleck here to see a explanation of how to solve the equation)

To find the y coordinates, substitute the above values into $ p(x) $

$$ \begin{aligned} \text{for } ~ x & = \color{blue}{ 0 } \Rightarrow p\left(0\right) = \color{orangered}{ 0 }\\[1 em] \text{for } ~ x & = \color{blue}{ 1.5136 } \Rightarrow p\left(1.5136\right) = \color{orangered}{ 35.4606 }\\[1 em] \text{for } ~ x & = \color{blue}{ -0.7136 } \Rightarrow p\left(-0.7136\right) = \color{orangered}{ -2.0974 }\end{aligned} $$

So the turning points are:

$$ \begin{matrix} \left( 0, 0 \right) & \left( 1.5136, 35.4606 \right) & \left( -0.7136, -2.0974 \right)\end{matrix} $$

STEP 5:Find the inflection points

The inflection points are located at zeroes of second derivative. The second derivative is $ p^{\prime \prime} (x) = -200x^3+120x^2+108x $.

The zeros of second derivative are

$$ \begin{matrix}x_1 = 0 & x_2 = 1.0937 & x_3 = -0.4937 \end{matrix} $$

Substitute the x values into $ p(x) $ to get y coordinates

$$ \begin{aligned} \text{for } ~ x & = \color{blue}{ 0 } \Rightarrow p\left(0\right) = \color{orangered}{ 0 }\\[1 em] \text{for } ~ x & = \color{blue}{ 1.0937 } \Rightarrow p\left(1.0937\right) = \color{orangered}{ 22.2092 }\\[1 em] \text{for } ~ x & = \color{blue}{ -0.4937 } \Rightarrow p\left(-0.4937\right) = \color{orangered}{ -1.2788 }\end{aligned} $$

So the inflection points are:

$$ \begin{matrix} \left( 0, 0 \right) & \left( 1.0937, 22.2092 \right) & \left( -0.4937, -1.2788 \right)\end{matrix} $$

Graphing Polynomials