Tap the blue points to see coordinates.
STEP 1:Find the x-intercepts
To find the x-intercepts solve, the equation $ \color{blue}{ x^3-6x-11 = 0 } $
The solution of this equation is:
$$ \begin{matrix}x = 3.0916 \end{matrix} $$(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) = x^3-6x-11 } $, so:
$$ \text{Y inercept} = p(0) = -11 $$STEP 3:Find the end behavior
The end behavior of a polynomial is the same as the end behavior of a leading term.
$$ \lim_{x \to -\infty} \left( x^3-6x-11 \right) = \lim_{x \to -\infty} x^3 = \color{blue}{ -\infty } $$The graph starts in the lower-left corner.
$$ \lim_{x \to \infty} \left( x^3-6x-11 \right) = \lim_{x \to \infty} x^3 = \color{blue}{ \infty } $$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) $:
$$ p^{\prime} (x) = 3x^2-6 $$The x coordinate of the turning points are located at the zeros of the first derivative
$$ p^{\prime} (x) = 0 $$ $$ \begin{matrix}x_1 = \sqrt{ 2 } & x_2 = -\sqrt{ 2 } \end{matrix} $$(cleck here to see a explanation of how to solve the equation)
To find the y coordinates, substitute the above values into $ p(x) $
$$ \begin{aligned} \text{for } ~ x & = \color{blue}{ \sqrt{ 2 } } \Rightarrow p\left(\sqrt{ 2 }\right) = \color{orangered}{ -16.6569 }\\[1 em] \text{for } ~ x & = \color{blue}{ -\sqrt{ 2 } } \Rightarrow p\left(-\sqrt{ 2 }\right) = \color{orangered}{ -5.3431 }\end{aligned} $$So the turning points are:
$$ \begin{matrix} \left( \sqrt{ 2 }, -16.6569 \right) & \left( -\sqrt{ 2 }, -5.3431 \right)\end{matrix} $$STEP 5:Find the inflection points
The inflection points are located at zeroes of second derivative. The second derivative is $ p^{\prime \prime} (x) = 6x $.
The zero of second derivative is
$$ \begin{matrix}x = 0 \end{matrix} $$Substitute the x value into $ p(x) $ to get y coordinates
$$ \begin{aligned} \text{for } ~ x & = \color{blue}{ 0 } \Rightarrow p\left(0\right) = \color{orangered}{ -11 }\end{aligned} $$So the inflection point is:
$$ \begin{matrix} \left( 0, -11 \right)\end{matrix} $$