Sketch the graph of the polynomial function $$ p(x) = 2x^4+16x^3+14x^2-72x-72 $$

Tap the blue points to see coordinates.

STEP 1:Find the x-intercepts

To find the x-intercepts solve, the equation $ \color{blue}{ 2x^4+16x^3+14x^2-72x-72 = 0 } $

The solutions of this equation are:

$$ \begin{matrix}x_1 = 2 & x_2 = -1 & x_3 = -3 & x_4 = -6 \end{matrix} $$

(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) = 2x^4+16x^3+14x^2-72x-72 } $, so:

$$ \text{Y inercept} = p(0) = -72 $$

STEP 3:Find the end behavior

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

$$ \lim_{x \to -\infty} \left( 2x^4+16x^3+14x^2-72x-72 \right) = \lim_{x \to -\infty} 2x^4 = \color{blue}{ \infty } $$

The graph starts in the upper-left corner.

$$ \lim_{x \to \infty} \left( 2x^4+16x^3+14x^2-72x-72 \right) = \lim_{x \to \infty} 2x^4 = \color{blue}{ \infty } $$

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) $:

$$ p^{\prime} (x) = 8x^3+48x^2+28x-72 $$

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

$$ p^{\prime} (x) = 0 $$ $$ \begin{matrix}x_1 = -2 & x_2 = 0.9155 & x_3 = -4.9155 \end{matrix} $$

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

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

$$ \begin{aligned} \text{for } ~ x & = \color{blue}{ -2 } \Rightarrow p\left(-2\right) = \color{orangered}{ 32 }\\[1 em] \text{for } ~ x & = \color{blue}{ 0.9155 } \Rightarrow p\left(0.9155\right) = \color{orangered}{ -112.5 }\\[1 em] \text{for } ~ x & = \color{blue}{ -4.9155 } \Rightarrow p\left(-4.9155\right) = \color{orangered}{ -112.5 }\end{aligned} $$

So the turning points are:

$$ \begin{matrix} \left( -2, 32 \right) & \left( 0.9155, -112.5 \right) & \left( -4.9155, -112.5 \right)\end{matrix} $$

STEP 5:Find the inflection points

The inflection points are located at zeroes of second derivative. The second derivative is $ p^{\prime \prime} (x) = 24x^2+96x+28 $.

The zeros of second derivative are

$$ \begin{matrix}x_1 = -0.3167 & x_2 = -3.6833 \end{matrix} $$

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

$$ \begin{aligned} \text{for } ~ x & = \color{blue}{ -0.3167 } \Rightarrow p\left(-0.3167\right) = \color{orangered}{ -48.2778 }\\[1 em] \text{for } ~ x & = \color{blue}{ -3.6833 } \Rightarrow p\left(-3.6833\right) = \color{orangered}{ -48.2778 }\end{aligned} $$

So the inflection points are:

$$ \begin{matrix} \left( -0.3167, -48.2778 \right) & \left( -3.6833, -48.2778 \right)\end{matrix} $$

Graphing Polynomials