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+3x^2-x-3 = 0 } $
The solutions of this equation are:
$$ \begin{matrix}x_1 = 1 & x_2 = -1 & x_3 = -3 \end{matrix} $$(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 $ x = 0 $ into $ \color{blue}{ p(x) = x^3+3x^2-x-3 } $, so:
$$ \text{Y inercept} = p(0) = -3 $$STEP 3:Find the end behavior
The end behavior of a polynomial is the same as the end behavior of a leading term.
$$ \lim_{x \to -\infty} \left( x^3+3x^2-x-3 \right) = \lim_{x \to -\infty} x^3 = \color{blue}{ -\infty } $$The graph starts in the lower-left corner.
$$ \lim_{x \to \infty} \left( x^3+3x^2-x-3 \right) = \lim_{x \to \infty} x^3 = \color{blue}{ \infty } $$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) $:
$$ p^{\prime} (x) = 3x^2+6x-1 $$The x coordinate of the turning points are located at the zeros of the first derivative
$$ p^{\prime} (x) = 0 $$ $$ \begin{matrix}x_1 = 0.1547 & x_2 = -2.1547 \end{matrix} $$(cleck here to see a explanation of how to solve the equation)
To find the y coordinates, substitute the above values into $ p(x) $
$$ \begin{aligned} \text{for } ~ x & = \color{blue}{ 0.1547 } \Rightarrow p\left(0.1547\right) = \color{orangered}{ -3.0792 }\\[1 em] \text{for } ~ x & = \color{blue}{ -2.1547 } \Rightarrow p\left(-2.1547\right) = \color{orangered}{ 3.0792 }\end{aligned} $$So the turning points are:
$$ \begin{matrix} \left( 0.1547, -3.0792 \right) & \left( -2.1547, 3.0792 \right)\end{matrix} $$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+6 $.
The zero of second derivative is
$$ \begin{matrix}x = -1 \end{matrix} $$Substitute the x value into $ p(x) $ to get y coordinates
$$ \begin{aligned} \text{for } ~ x & = \color{blue}{ -1 } \Rightarrow p\left(-1\right) = \color{orangered}{ 0 }\end{aligned} $$So the inflection point is:
$$ \begin{matrix} \left( -1, 0 \right)\end{matrix} $$