Sketch the graph of the polynomial function $$ p(x) = -x^4+12x^3-31x^2-48x+140 $$

Tap the blue points to see coordinates.

STEP 1:Find the x-intercepts

To find the x-intercepts solve, the equation $ \color{blue}{ -x^4+12x^3-31x^2-48x+140 = 0 } $

The solutions of this equation are:

$$ \begin{matrix}x_1 = 2 & x_2 = 5 & x_3 = 7 & x_4 = -2 \end{matrix} $$

(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) = -x^4+12x^3-31x^2-48x+140 } $, so:

$$ \text{Y inercept} = p(0) = 140 $$

STEP 3:Find the end behavior

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

$$ \lim_{x \to -\infty} \left( -x^4+12x^3-31x^2-48x+140 \right) = \lim_{x \to -\infty} -x^4 = \color{blue}{ -\infty } $$

The graph starts in the lower-left corner.

$$ \lim_{x \to \infty} \left( -x^4+12x^3-31x^2-48x+140 \right) = \lim_{x \to \infty} -x^4 = \color{blue}{ -\infty } $$

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) $:

$$ p^{\prime} (x) = -4x^3+36x^2-62x-48 $$

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

$$ p^{\prime} (x) = 0 $$ $$ \begin{matrix}x_1 = -0.5721 & x_2 = 3.3967 & x_3 = 6.1754 \end{matrix} $$

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

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

$$ \begin{aligned} \text{for } ~ x & = \color{blue}{ -0.5721 } \Rightarrow p\left(-0.5721\right) = \color{orangered}{ 154.9605 }\\[1 em] \text{for } ~ x & = \color{blue}{ 3.3967 } \Rightarrow p\left(3.3967\right) = \color{orangered}{ -43.5458 }\\[1 em] \text{for } ~ x & = \color{blue}{ 6.1754 } \Rightarrow p\left(6.1754\right) = \color{orangered}{ 33.0854 }\end{aligned} $$

So the turning points are:

$$ \begin{matrix} \left( -0.5721, 154.9605 \right) & \left( 3.3967, -43.5458 \right) & \left( 6.1754, 33.0854 \right)\end{matrix} $$

STEP 5:Find the inflection points

The inflection points are located at zeroes of second derivative. The second derivative is $ p^{\prime \prime} (x) = -12x^2+72x-62 $.

The zeros of second derivative are

$$ \begin{matrix}x_1 = 4.9579 & x_2 = 1.0421 \end{matrix} $$

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

$$ \begin{aligned} \text{for } ~ x & = \color{blue}{ 4.9579 } \Rightarrow p\left(4.9579\right) = \color{orangered}{ -1.7698 }\\[1 em] \text{for } ~ x & = \color{blue}{ 1.0421 } \Rightarrow p\left(1.0421\right) = \color{orangered}{ 68.7142 }\end{aligned} $$

So the inflection points are:

$$ \begin{matrix} \left( 4.9579, -1.7698 \right) & \left( 1.0421, 68.7142 \right)\end{matrix} $$

Graphing Polynomials