Sketch the graph of the polynomial function $$ p(x) = 2x^4+5x^3-10x^2-15x+18 $$

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+5x^3-10x^2-15x+18 = 0 } $

The solutions of this equation are:

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

(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) = 2x^4+5x^3-10x^2-15x+18 } $, so:

$$ \text{Y inercept} = p(0) = 18 $$

STEP 3:Find the end behavior

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

$$ \lim_{x \to -\infty} \left( 2x^4+5x^3-10x^2-15x+18 \right) = \lim_{x \to -\infty} 2x^4 = \color{blue}{ \infty } $$

The graph starts in the upper-left corner.

$$ \lim_{x \to \infty} \left( 2x^4+5x^3-10x^2-15x+18 \right) = \lim_{x \to \infty} 2x^4 = \color{blue}{ \infty } $$

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) $:

$$ p^{\prime} (x) = 8x^3+15x^2-20x-15 $$

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

$$ p^{\prime} (x) = 0 $$ $$ \begin{matrix}x_1 = -0.5771 & x_2 = 1.2668 & x_3 = -2.5647 \end{matrix} $$

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

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

$$ \begin{aligned} \text{for } ~ x & = \color{blue}{ -0.5771 } \Rightarrow p\left(-0.5771\right) = \color{orangered}{ 22.5869 }\\[1 em] \text{for } ~ x & = \color{blue}{ 1.2668 } \Rightarrow p\left(1.2668\right) = \color{orangered}{ -1.7345 }\\[1 em] \text{for } ~ x & = \color{blue}{ -2.5647 } \Rightarrow p\left(-2.5647\right) = \color{orangered}{ -7.1234 }\end{aligned} $$

So the turning points are:

$$ \begin{matrix} \left( -0.5771, 22.5869 \right) & \left( 1.2668, -1.7345 \right) & \left( -2.5647, -7.1234 \right)\end{matrix} $$

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+30x-20 $.

The zeros of second derivative are

$$ \begin{matrix}x_1 = 0.4813 & x_2 = -1.7313 \end{matrix} $$

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

$$ \begin{aligned} \text{for } ~ x & = \color{blue}{ 0.4813 } \Rightarrow p\left(0.4813\right) = \color{orangered}{ 9.1283 }\\[1 em] \text{for } ~ x & = \color{blue}{ -1.7313 } \Rightarrow p\left(-1.7313\right) = \color{orangered}{ 6.0167 }\end{aligned} $$

So the inflection points are:

$$ \begin{matrix} \left( 0.4813, 9.1283 \right) & \left( -1.7313, 6.0167 \right)\end{matrix} $$

Graphing Polynomials