STEP 1:Find the x-intercepts

To find the x-intercepts solve, the equation $ \color{blue}{ 3x^5-2x^4-6x^3 = 0 } $

The solutions of this equation are:

$$ \begin{matrix}x_1 = 0 & x_2 = 1.7863 & x_3 = -1.1196 \end{matrix} $$

(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) = 3x^5-2x^4-6x^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.

$$ \lim_{x \to -\infty} \left( 3x^5-2x^4-6x^3 \right) = \lim_{x \to -\infty} 3x^5 = \color{blue}{ -\infty } $$

The graph starts in the lower-left corner.

$$ \lim_{x \to \infty} \left( 3x^5-2x^4-6x^3 \right) = \lim_{x \to \infty} 3x^5 = \color{blue}{ \infty } $$

The graph ends in the upper-right corner.

STEP 4:Find the turning points

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

$$ p^{\prime} (x) = 15x^4-8x^3-18x^2 $$

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

$$ p^{\prime} (x) = 0 $$ $$ \begin{matrix}x_1 = 0 & x_2 = 1.3941 & x_3 = -0.8608 \end{matrix} $$

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

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

$$ \begin{aligned} \text{for } ~ x & = \color{blue}{ 0 } \Rightarrow p\left(0\right) = \color{orangered}{ 0 }\\[1 em] \text{for } ~ x & = \color{blue}{ 1.3941 } \Rightarrow p\left(1.3941\right) = \color{orangered}{ -8.0136 }\\[1 em] \text{for } ~ x & = \color{blue}{ -0.8608 } \Rightarrow p\left(-0.8608\right) = \color{orangered}{ 1.311 }\end{aligned} $$

So the turning points are:

$$ \begin{matrix} \left( 0, 0 \right) & \left( 1.3941, -8.0136 \right) & \left( -0.8608, 1.311 \right)\end{matrix} $$

STEP 5:Find the inflection points

The inflection points are located at zeroes of second derivative. The second derivative is $ p^{\prime \prime} (x) = 60x^3-24x^2-36x $.

The zeros of second derivative are

$$ \begin{matrix}x_1 = 0 & x_2 = 1 & x_3 = -\dfrac{ 3 }{ 5 } \end{matrix} $$

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

$$ \begin{aligned} \text{for } ~ x & = \color{blue}{ 0 } \Rightarrow p\left(0\right) = \color{orangered}{ 0 }\\[1 em] \text{for } ~ x & = \color{blue}{ 1 } \Rightarrow p\left(1\right) = \color{orangered}{ -5 }\\[1 em] \text{for } ~ x & = \color{blue}{ -\frac{ 3 }{ 5 } } \Rightarrow p\left(-\frac{ 3 }{ 5 }\right) = \color{orangered}{ \frac{ 2511 }{ 3125 } }\end{aligned} $$

So the inflection points are:

$$ \begin{matrix} \left( 0, 0 \right) & \left( 1, -5 \right) & \left( -\dfrac{ 3 }{ 5 }, \dfrac{ 2511 }{ 3125 } \right)\end{matrix} $$

Graphing Polynomials