Tap the blue points to see coordinates.
STEP 1:Find the x-intercepts
To find the x-intercepts solve, the equation $ \color{blue}{ x^5-x^4-5x^3+x^2+8x+4 = 0 } $
The solutions of this equation are:
$$ \begin{matrix}x_1 = 2 & x_2 = -1 & x_3 = 2 & x_4 = -1 & x_5 = -1 \end{matrix} $$(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^5-x^4-5x^3+x^2+8x+4 } $, so:
$$ \text{Y inercept} = p(0) = 4 $$STEP 3:Find the end behavior
The end behavior of a polynomial is the same as the end behavior of a leading term.
$$ \lim_{x \to -\infty} \left( x^5-x^4-5x^3+x^2+8x+4 \right) = \lim_{x \to -\infty} x^5 = \color{blue}{ -\infty } $$The graph starts in the lower-left corner.
$$ \lim_{x \to \infty} \left( x^5-x^4-5x^3+x^2+8x+4 \right) = \lim_{x \to \infty} x^5 = \color{blue}{ \infty } $$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) $:
$$ p^{\prime} (x) = 5x^4-4x^3-15x^2+2x+8 $$The x coordinate of the turning points are located at the zeros of the first derivative
$$ p^{\prime} (x) = 0 $$ $$ \begin{matrix}x_1 = 2 & x_2 = -1 & x_3 = \dfrac{ 4 }{ 5 } & x_4 = -1 \end{matrix} $$(cleck here to see a explanation of how to solve the equation)
To find the y coordinates, substitute the above values into $ p(x) $
$$ \begin{aligned} \text{for } ~ x & = \color{blue}{ 2 } \Rightarrow p\left(2\right) = \color{orangered}{ 0 }\\[1 em] \text{for } ~ x & = \color{blue}{ -1 } \Rightarrow p\left(-1\right) = \color{orangered}{ 0 }\\[1 em] \text{for } ~ x & = \color{blue}{ \frac{ 4 }{ 5 } } \Rightarrow p\left(\frac{ 4 }{ 5 }\right) = \color{orangered}{ \frac{ 26244 }{ 3125 } }\\[1 em] \text{for } ~ x & = \color{blue}{ -1 } \Rightarrow p\left(-1\right) = \color{orangered}{ 0 }\end{aligned} $$So the turning points are:
$$ \begin{matrix} \left( 2, 0 \right) & \left( -1, 0 \right) & \left( \dfrac{ 4 }{ 5 }, \dfrac{ 26244 }{ 3125 } \right) & \left( -1, 0 \right)\end{matrix} $$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^3-12x^2-30x+2 $.
The zeros of second derivative are
$$ \begin{matrix}x_1 = -1 & x_2 = 1.5348 & x_3 = 0.0652 \end{matrix} $$Substitute the x values into $ p(x) $ to get y coordinates
$$ \begin{aligned} \text{for } ~ x & = \color{blue}{ -1 } \Rightarrow p\left(-1\right) = \color{orangered}{ 0 }\\[1 em] \text{for } ~ x & = \color{blue}{ 1.5348 } \Rightarrow p\left(1.5348\right) = \color{orangered}{ 3.5241 }\\[1 em] \text{for } ~ x & = \color{blue}{ 0.0652 } \Rightarrow p\left(0.0652\right) = \color{orangered}{ 4.5241 }\end{aligned} $$So the inflection points are:
$$ \begin{matrix} \left( -1, 0 \right) & \left( 1.5348, 3.5241 \right) & \left( 0.0652, 4.5241 \right)\end{matrix} $$