Sketch the graph of the polynomial function $$ p(x) = \frac{1}{12}x^4-\frac{1}{3}x^3-\frac{7}{12}x^2+\frac{5}{6}x $$

Tap the blue points to see coordinates.

STEP 1:Find the x-intercepts

To find the x-intercepts solve, the equation $ \color{blue}{ \frac{1}{12}x^4-\frac{1}{3}x^3-\frac{7}{12}x^2+\frac{5}{6}x = 0 } $

The solutions of this equation are:

$$ \begin{matrix}x_1 = 0 & x_2 = 1 & x_3 = 5 & x_4 = -2 \end{matrix} $$

(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) = \frac{1}{12}x^4-\frac{1}{3}x^3-\frac{7}{12}x^2+\frac{5}{6}x } $, 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.

$$ \lim_{x \to -\infty} \left( \frac{1}{12}x^4-\frac{1}{3}x^3-\frac{7}{12}x^2+\frac{5}{6}x \right) = \lim_{x \to -\infty} \frac{1}{12}x^4 = \color{blue}{ \infty } $$

The graph starts in the upper-left corner.

$$ \lim_{x \to \infty} \left( \frac{1}{12}x^4-\frac{1}{3}x^3-\frac{7}{12}x^2+\frac{5}{6}x \right) = \lim_{x \to \infty} \frac{1}{12}x^4 = \color{blue}{ \infty } $$

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) $:

$$ p^{\prime} (x) = \frac{1}{3}x^3-x^2-\frac{7}{6}x+\frac{5}{6} $$

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

$$ p^{\prime} (x) = 0 $$ $$ \begin{matrix}x_1 = 0.5216 & x_2 = -1.2764 & x_3 = 3.7548 \end{matrix} $$

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

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

$$ \begin{aligned} \text{for } ~ x & = \color{blue}{ 0.5216 } \Rightarrow p\left(0.5216\right) = \color{orangered}{ 0.2348 }\\[1 em] \text{for } ~ x & = \color{blue}{ -1.2764 } \Rightarrow p\left(-1.2764\right) = \color{orangered}{ -1.0997 }\\[1 em] \text{for } ~ x & = \color{blue}{ 3.7548 } \Rightarrow p\left(3.7548\right) = \color{orangered}{ -6.1768 }\end{aligned} $$

So the turning points are:

$$ \begin{matrix} \left( 0.5216, 0.2348 \right) & \left( -1.2764, -1.0997 \right) & \left( 3.7548, -6.1768 \right)\end{matrix} $$

STEP 5:Find the inflection points

The inflection points are located at zeroes of second derivative. The second derivative is $ p^{\prime \prime} (x) = x^2-2x-\frac{7}{6} $.

The zeros of second derivative are

$$ \begin{matrix}x_1 = 2.472 & x_2 = -0.472 \end{matrix} $$

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

$$ \begin{aligned} \text{for } ~ x & = \color{blue}{ 2.472 } \Rightarrow p\left(2.472\right) = \color{orangered}{ -3.428 }\\[1 em] \text{for } ~ x & = \color{blue}{ -0.472 } \Rightarrow p\left(-0.472\right) = \color{orangered}{ -0.4841 }\end{aligned} $$

So the inflection points are:

$$ \begin{matrix} \left( 2.472, -3.428 \right) & \left( -0.472, -0.4841 \right)\end{matrix} $$

Graphing Polynomials