Tap the blue points to see coordinates.
STEP 1:Find the x-intercepts
To find the x-intercepts solve, the equation $ \color{blue}{ 12x^3-36x-x^5 = 0 } $
The solutions of this equation are:
$$ \begin{matrix}x_1 = 0 & x_2 = 2.4495 & x_3 = 2.4495 & x_4 = -2.4495 & x_5 = -2.4495 \end{matrix} $$(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) = 12x^3-36x-x^5 } $, 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.
$$ \lim_{x \to -\infty} \left( 12x^3-36x-x^5 \right) = \lim_{x \to -\infty} 12x^3 = \color{blue}{ -\infty } $$The graph starts in the lower-left corner.
$$ \lim_{x \to \infty} \left( 12x^3-36x-x^5 \right) = \lim_{x \to \infty} 12x^3 = \color{blue}{ \infty } $$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) $:
$$ p^{\prime} (x) = -5x^4+36x^2-36 $$The x coordinate of the turning points are located at the zeros of the first derivative
$$ p^{\prime} (x) = 0 $$ $$ \begin{matrix}x_1 = -1.0955 & x_2 = 1.0955 & x_3 = 2.4495 & x_4 = -2.4495 \end{matrix} $$(cleck here to see a explanation of how to solve the equation)
To find the y coordinates, substitute the above values into $ p(x) $
$$ \begin{aligned} \text{for } ~ x & = \color{blue}{ -1.0955 } \Rightarrow p\left(-1.0955\right) = \color{orangered}{ 25.2391 }\\[1 em] \text{for } ~ x & = \color{blue}{ 1.0955 } \Rightarrow p\left(1.0955\right) = \color{orangered}{ -25.2391 }\\[1 em] \text{for } ~ x & = \color{blue}{ 2.4495 } \Rightarrow p\left(2.4495\right) = \color{orangered}{ -0 }\\[1 em] \text{for } ~ x & = \color{blue}{ -2.4495 } \Rightarrow p\left(-2.4495\right) = \color{orangered}{ 0 }\end{aligned} $$So the turning points are:
$$ \begin{matrix} \left( -1.0955, 25.2391 \right) & \left( 1.0955, -25.2391 \right) & \left( 2.4495, -0 \right) & \left( -2.4495, 0 \right)\end{matrix} $$STEP 5:Find the inflection points
The inflection points are located at zeroes of second derivative. The second derivative is $ p^{\prime \prime} (x) = -20x^3+72x $.
The zeros of second derivative are
$$ \begin{matrix}x_1 = 0 & x_2 = \dfrac{ 3 \sqrt{ 10}}{ 5 } & x_3 = -3 \dfrac{\sqrt{ 10 }}{ 5 } \end{matrix} $$Substitute the x values into $ p(x) $ to get y coordinates
$$ \begin{aligned} \text{for } ~ x & = \color{blue}{ 0 } \Rightarrow p\left(0\right) = \color{orangered}{ 0 }\\[1 em] \text{for } ~ x & = \color{blue}{ \frac{ 3 \sqrt{ 10}}{ 5 } } \Rightarrow p\left(\frac{ 3 \sqrt{ 10}}{ 5 }\right) = \color{orangered}{ -432 \frac{\sqrt{ 10 }}{ 125 } }\\[1 em] \text{for } ~ x & = \color{blue}{ -3 \frac{\sqrt{ 10 }}{ 5 } } \Rightarrow p\left(-3 \frac{\sqrt{ 10 }}{ 5 }\right) = \color{orangered}{ \frac{ 432 \sqrt{ 10}}{ 125 } }\end{aligned} $$So the inflection points are:
$$ \begin{matrix} \left( 0, 0 \right) & \left( \dfrac{ 3 \sqrt{ 10}}{ 5 }, -432 \dfrac{\sqrt{ 10 }}{ 125 } \right) & \left( -3 \dfrac{\sqrt{ 10 }}{ 5 }, \dfrac{ 432 \sqrt{ 10}}{ 125 } \right)\end{matrix} $$