Sketch the graph of the polynomial function $$ p(x) = 2x^5-9x^4+15x^3-11x^2+3x $$

Tap the blue points to see coordinates.

STEP 1:Find the x-intercepts

To find the x-intercepts solve, the equation $ \color{blue}{ 2x^5-9x^4+15x^3-11x^2+3x = 0 } $

The solutions of this equation are:

$$ \begin{matrix}x_1 = 0 & x_2 = 1 & x_3 = \dfrac{ 3 }{ 2 } & x_4 = 1 & x_5 = 1 \end{matrix} $$

(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^5-9x^4+15x^3-11x^2+3x } $, 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.

$$ \lim_{x \to -\infty} \left( 2x^5-9x^4+15x^3-11x^2+3x \right) = \lim_{x \to -\infty} 2x^5 = \color{blue}{ -\infty } $$

The graph starts in the lower-left corner.

$$ \lim_{x \to \infty} \left( 2x^5-9x^4+15x^3-11x^2+3x \right) = \lim_{x \to \infty} 2x^5 = \color{blue}{ \infty } $$

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) $:

$$ p^{\prime} (x) = 10x^4-36x^3+45x^2-22x+3 $$

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

$$ p^{\prime} (x) = 0 $$ $$ \begin{matrix}x_1 = 1 & x_2 = 1 & x_3 = 1.3831 & x_4 = 0.2169 \end{matrix} $$

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

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

$$ \begin{aligned} \text{for } ~ x & = \color{blue}{ 1 } \Rightarrow p\left(1\right) = \color{orangered}{ 0 }\\[1 em] \text{for } ~ x & = \color{blue}{ 1 } \Rightarrow p\left(1\right) = \color{orangered}{ 0 }\\[1 em] \text{for } ~ x & = \color{blue}{ 1.3831 } \Rightarrow p\left(1.3831\right) = \color{orangered}{ -0.0182 }\\[1 em] \text{for } ~ x & = \color{blue}{ 0.2169 } \Rightarrow p\left(0.2169\right) = \color{orangered}{ 0.2673 }\end{aligned} $$

So the turning points are:

$$ \begin{matrix} \left( 1, 0 \right) & \left( 1, 0 \right) & \left( 1.3831, -0.0182 \right) & \left( 0.2169, 0.2673 \right)\end{matrix} $$

STEP 5:Find the inflection points

The inflection points are located at zeroes of second derivative. The second derivative is $ p^{\prime \prime} (x) = 40x^3-108x^2+90x-22 $.

The zeros of second derivative are

$$ \begin{matrix}x_1 = 1 & x_2 = 1.2653 & x_3 = 0.4347 \end{matrix} $$

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

$$ \begin{aligned} \text{for } ~ x & = \color{blue}{ 1 } \Rightarrow p\left(1\right) = \color{orangered}{ 0 }\\[1 em] \text{for } ~ x & = \color{blue}{ 1.2653 } \Rightarrow p\left(1.2653\right) = \color{orangered}{ -0.0111 }\\[1 em] \text{for } ~ x & = \color{blue}{ 0.4347 } \Rightarrow p\left(0.4347\right) = \color{orangered}{ 0.1673 }\end{aligned} $$

So the inflection points are:

$$ \begin{matrix} \left( 1, 0 \right) & \left( 1.2653, -0.0111 \right) & \left( 0.4347, 0.1673 \right)\end{matrix} $$

Graphing Polynomials