Sketch the graph of the polynomial function $$ p(x) = 3x^5-4x^3+81x^2-108 $$

Tap the blue points to see coordinates.

STEP 1:Find the x-intercepts

To find the x-intercepts solve, the equation $ \color{blue}{ 3x^5-4x^3+81x^2-108 = 0 } $

The solutions of this equation are:

$$ \begin{matrix}x_1 = -3 & x_2 = 1.1547 & x_3 = -1.1547 \end{matrix} $$

(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^5-4x^3+81x^2-108 } $, so:

$$ \text{Y inercept} = p(0) = -108 $$

STEP 3:Find the end behavior

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

$$ \lim_{x \to -\infty} \left( 3x^5-4x^3+81x^2-108 \right) = \lim_{x \to -\infty} 3x^5 = \color{blue}{ -\infty } $$

The graph starts in the lower-left corner.

$$ \lim_{x \to \infty} \left( 3x^5-4x^3+81x^2-108 \right) = \lim_{x \to \infty} 3x^5 = \color{blue}{ \infty } $$

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) $:

$$ p^{\prime} (x) = 15x^4-12x^2+162x $$

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

$$ p^{\prime} (x) = 0 $$ $$ \begin{matrix}x_1 = 0 & x_2 = -2.331 \end{matrix} $$

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

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

$$ \begin{aligned} \text{for } ~ x & = \color{blue}{ 0 } \Rightarrow p\left(0\right) = \color{orangered}{ -108 }\\[1 em] \text{for } ~ x & = \color{blue}{ -2.331 } \Rightarrow p\left(-2.331\right) = \color{orangered}{ 176.3225 }\end{aligned} $$

So the turning points are:

$$ \begin{matrix} \left( 0, -108 \right) & \left( -2.331, 176.3225 \right)\end{matrix} $$

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+162 $.

The zero of second derivative is

$$ \begin{matrix}x = -1.4881 \end{matrix} $$

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

$$ \begin{aligned} \text{for } ~ x & = \color{blue}{ -1.4881 } \Rightarrow p\left(-1.4881\right) = \color{orangered}{ 62.6573 }\end{aligned} $$

So the inflection point is:

$$ \begin{matrix} \left( -1.4881, 62.6573 \right)\end{matrix} $$

Graphing Polynomials