Sketch the graph of the polynomial function $$ p(x) = 3x^5+2x^4-87x^3-58x^2+300x+200 $$

Tap the blue points to see coordinates.

STEP 1:Find the x-intercepts

To find the x-intercepts solve, the equation $ \color{blue}{ 3x^5+2x^4-87x^3-58x^2+300x+200 = 0 } $

The solutions of this equation are:

$$ \begin{matrix}x_1 = 2 & x_2 = 5 & x_3 = -2 & x_4 = -5 & x_5 = -\dfrac{ 2 }{ 3 } \end{matrix} $$

(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) = 3x^5+2x^4-87x^3-58x^2+300x+200 } $, so:

$$ \text{Y inercept} = p(0) = 200 $$

STEP 3:Find the end behavior

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

$$ \lim_{x \to -\infty} \left( 3x^5+2x^4-87x^3-58x^2+300x+200 \right) = \lim_{x \to -\infty} 3x^5 = \color{blue}{ -\infty } $$

The graph starts in the lower-left corner.

$$ \lim_{x \to \infty} \left( 3x^5+2x^4-87x^3-58x^2+300x+200 \right) = \lim_{x \to \infty} 3x^5 = \color{blue}{ \infty } $$

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) $:

$$ p^{\prime} (x) = 15x^4+8x^3-261x^2-116x+300 $$

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

$$ p^{\prime} (x) = 0 $$ $$ \begin{matrix}x_1 = 0.8997 & x_2 = -1.3727 & x_3 = 3.994 & x_4 = -4.0544 \end{matrix} $$

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

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

$$ \begin{aligned} \text{for } ~ x & = \color{blue}{ 0.8997 } \Rightarrow p\left(0.8997\right) = \color{orangered}{ 362.6807 }\\[1 em] \text{for } ~ x & = \color{blue}{ -1.3727 } \Rightarrow p\left(-1.3727\right) = \color{orangered}{ -103.5873 }\\[1 em] \text{for } ~ x & = \color{blue}{ 3.994 } \Rightarrow p\left(3.994\right) = \color{orangered}{ -1512.0357 }\\[1 em] \text{for } ~ x & = \color{blue}{ -4.0544 } \Rightarrow p\left(-4.0544\right) = \color{orangered}{ 1082.3151 }\end{aligned} $$

So the turning points are:

$$ \begin{matrix} \left( 0.8997, 362.6807 \right) & \left( -1.3727, -103.5873 \right) & \left( 3.994, -1512.0357 \right) & \left( -4.0544, 1082.3151 \right)\end{matrix} $$

STEP 5:Find the inflection points

The inflection points are located at zeroes of second derivative. The second derivative is $ p^{\prime \prime} (x) = 60x^3+24x^2-522x-116 $.

The zeros of second derivative are

$$ \begin{matrix}x_1 = -0.2212 & x_2 = 2.8682 & x_3 = -3.047 \end{matrix} $$

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

$$ \begin{aligned} \text{for } ~ x & = \color{blue}{ -0.2212 } \Rightarrow p\left(-0.2212\right) = \color{orangered}{ 131.7407 }\\[1 em] \text{for } ~ x & = \color{blue}{ 2.8682 } \Rightarrow p\left(2.8682\right) = \color{orangered}{ -751.8328 }\\[1 em] \text{for } ~ x & = \color{blue}{ -3.047 } \Rightarrow p\left(-3.047\right) = \color{orangered}{ 593.0422 }\end{aligned} $$

So the inflection points are:

$$ \begin{matrix} \left( -0.2212, 131.7407 \right) & \left( 2.8682, -751.8328 \right) & \left( -3.047, 593.0422 \right)\end{matrix} $$

Graphing Polynomials