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+3x+2 = 0 } $
The solution of this equation is:
$$ \begin{matrix}x = -0.4423 \end{matrix} $$(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+3x+2 } $, so:
$$ \text{Y inercept} = p(0) = 2 $$STEP 3:Find the end behavior
The end behavior of a polynomial is the same as the end behavior of a leading term.
$$ \lim_{x \to -\infty} \left( x^3-3x^2+3x+2 \right) = \lim_{x \to -\infty} x^3 = \color{blue}{ -\infty } $$The graph starts in the lower-left corner.
$$ \lim_{x \to \infty} \left( x^3-3x^2+3x+2 \right) = \lim_{x \to \infty} x^3 = \color{blue}{ \infty } $$The graph ends in the upper-right corner.
STEP 4:Find the turning points
To determine the turning point, we need to find the first derivative of $ p(x) $:
$$ p^{\prime} (x) = 3x^2-6x+3 $$The x coordinate of the turning point is located at the zeros of the first derivative
$$ p^{\prime} (x) = 0 $$ $$ \begin{matrix}x = 1 \end{matrix} $$(cleck here to see a explanation of how to solve the equation)
To find the y coordinate, substitute the above value into $ p(x) $
$$ \begin{aligned} \text{for } ~ x & = \color{blue}{ 1 } \Rightarrow p\left(1\right) = \color{orangered}{ 3 }\end{aligned} $$So the turning point is:
$$ \begin{matrix} \left( 1, 3 \right)\end{matrix} $$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
$$ \begin{aligned} \text{for } ~ x & = \color{blue}{ 1 } \Rightarrow p\left(1\right) = \color{orangered}{ 3 }\end{aligned} $$So the inflection point is:
$$ \begin{matrix} \left( 1, 3 \right)\end{matrix} $$