Sketch the graph of the polynomial function $$ p(x) = x^4+3x^3-13x^2-27x+36 $$

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+3x^3-13x^2-27x+36 = 0 } $

The solutions of this equation are:

$$ \begin{matrix}x_1 = 1 & x_2 = 3 & x_3 = -3 & x_4 = -4 \end{matrix} $$

(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+3x^3-13x^2-27x+36 } $, so:

$$ \text{Y inercept} = p(0) = 36 $$

STEP 3:Find the end behavior

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

$$ \lim_{x \to -\infty} \left( x^4+3x^3-13x^2-27x+36 \right) = \lim_{x \to -\infty} x^4 = \color{blue}{ \infty } $$

The graph starts in the upper-left corner.

$$ \lim_{x \to \infty} \left( x^4+3x^3-13x^2-27x+36 \right) = \lim_{x \to \infty} x^4 = \color{blue}{ \infty } $$

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) $:

$$ p^{\prime} (x) = 4x^3+9x^2-26x-27 $$

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

$$ p^{\prime} (x) = 0 $$ $$ \begin{matrix}x_1 = -0.8762 & x_2 = 2.1724 & x_3 = -3.5462 \end{matrix} $$

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

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

$$ \begin{aligned} \text{for } ~ x & = \color{blue}{ -0.8762 } \Rightarrow p\left(-0.8762\right) = \color{orangered}{ 48.2483 }\\[1 em] \text{for } ~ x & = \color{blue}{ 2.1724 } \Rightarrow p\left(2.1724\right) = \color{orangered}{ -30.9772 }\\[1 em] \text{for } ~ x & = \color{blue}{ -3.5462 } \Rightarrow p\left(-3.5462\right) = \color{orangered}{ -7.3766 }\end{aligned} $$

So the turning points are:

$$ \begin{matrix} \left( -0.8762, 48.2483 \right) & \left( 2.1724, -30.9772 \right) & \left( -3.5462, -7.3766 \right)\end{matrix} $$

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+18x-26 $.

The zeros of second derivative are

$$ \begin{matrix}x_1 = 0.902 & x_2 = -2.402 \end{matrix} $$

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

$$ \begin{aligned} \text{for } ~ x & = \color{blue}{ 0.902 } \Rightarrow p\left(0.902\right) = \color{orangered}{ 3.9319 }\\[1 em] \text{for } ~ x & = \color{blue}{ -2.402 } \Rightarrow p\left(-2.402\right) = \color{orangered}{ 17.5611 }\end{aligned} $$

So the inflection points are:

$$ \begin{matrix} \left( 0.902, 3.9319 \right) & \left( -2.402, 17.5611 \right)\end{matrix} $$

Graphing Polynomials