Sketch the graph of the polynomial function $$ p(r) = -r^4-r^3+6r^2 $$

Tap the blue points to see coordinates.

STEP 1:Find the x-intercepts

To find the x-intercepts solve, the equation $ \color{blue}{ -r^4-r^3+6r^2 = 0 } $

The solutions of this equation are:

$$ \begin{matrix}r_1 = 0 & r_2 = 2 & r_3 = -3 \end{matrix} $$

(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 $ r = 0 $ into $ \color{blue}{ p(r) = -r^4-r^3+6r^2 } $, 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.

$$ \lim_{x \to -\infty} \left( -r^4-r^3+6r^2 \right) = \lim_{x \to -\infty} -r^4 = \color{blue}{ -\infty } $$

The graph starts in the lower-left corner.

$$ \lim_{x \to \infty} \left( -r^4-r^3+6r^2 \right) = \lim_{x \to \infty} -r^4 = \color{blue}{ -\infty } $$

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(r) $:

$$ p^{\prime} (x) = -4r^3-3r^2+12r $$

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

$$ p^{\prime} (r) = 0 $$ $$ \begin{matrix}r_1 = 0 & r_2 = 1.3972 & r_3 = -2.1472 \end{matrix} $$

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

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

$$ \begin{aligned} \text{for } ~ r & = \color{blue}{ 0 } \Rightarrow p\left(0\right) = \color{orangered}{ 0 }\\[1 em] \text{for } ~ r & = \color{blue}{ 1.3972 } \Rightarrow p\left(1.3972\right) = \color{orangered}{ 5.1745 }\\[1 em] \text{for } ~ r & = \color{blue}{ -2.1472 } \Rightarrow p\left(-2.1472\right) = \color{orangered}{ 16.306 }\end{aligned} $$

So the turning points are:

$$ \begin{matrix} \left( 0, 0 \right) & \left( 1.3972, 5.1745 \right) & \left( -2.1472, 16.306 \right)\end{matrix} $$

STEP 5:Find the inflection points

The inflection points are located at zeroes of second derivative. The second derivative is $ p^{\prime \prime} (x) = -12r^2-6r+12 $.

The zeros of second derivative are

$$ \begin{matrix}r_1 = 0.7808 & r_2 = -1.2808 \end{matrix} $$

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

$$ \begin{aligned} \text{for } ~ r & = \color{blue}{ 0.7808 } \Rightarrow p\left(0.7808\right) = \color{orangered}{ 2.8101 }\\[1 em] \text{for } ~ r & = \color{blue}{ -1.2808 } \Rightarrow p\left(-1.2808\right) = \color{orangered}{ 9.2524 }\end{aligned} $$

So the inflection points are:

$$ \begin{matrix} \left( 0.7808, 2.8101 \right) & \left( -1.2808, 9.2524 \right)\end{matrix} $$

Graphing Polynomials