STEP 1:Find the x-intercepts

To find the x-intercepts solve, the equation $ \color{blue}{ x^5-5x^4+5x^3+5x^2-6x-1 = 0 } $

The solutions of this equation are:

$$ \begin{matrix}x_1 = -0.151 & x_2 = -0.9541 & x_3 = 1.2758 & x_4 = 1.7907 & x_5 = 3.0385 \end{matrix} $$

(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^5-5x^4+5x^3+5x^2-6x-1 } $, so:

$$ \text{Y inercept} = p(0) = -1 $$

STEP 3:Find the end behavior

The end behavior of a polynomial is the same as the end behavior of a leading term.

$$ \lim_{x \to -\infty} \left( x^5-5x^4+5x^3+5x^2-6x-1 \right) = \lim_{x \to -\infty} x^5 = \color{blue}{ -\infty } $$

The graph starts in the lower-left corner.

$$ \lim_{x \to \infty} \left( x^5-5x^4+5x^3+5x^2-6x-1 \right) = \lim_{x \to \infty} x^5 = \color{blue}{ \infty } $$

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) $:

$$ p^{\prime} (x) = 5x^4-20x^3+15x^2+10x-6 $$

The x coordinate of the turning points are located at the zeros of the first derivative

$$ p^{\prime} (x) = 0 $$ $$ \begin{matrix}x_1 = 0.4561 & x_2 = -0.6444 & x_3 = 1.5439 & x_4 = 2.6444 \end{matrix} $$

(cleck here to see a explanation of how to solve the equation)

To find the y coordinates, substitute the above values into $ p(x) $

$$ \begin{aligned} \text{for } ~ x & = \color{blue}{ 0.4561 } \Rightarrow p\left(0.4561\right) = \color{orangered}{ -2.4187 }\\[1 em] \text{for } ~ x & = \color{blue}{ -0.6444 } \Rightarrow p\left(-0.6444\right) = \color{orangered}{ 2.6314 }\\[1 em] \text{for } ~ x & = \color{blue}{ 1.5439 } \Rightarrow p\left(1.5439\right) = \color{orangered}{ 0.4187 }\\[1 em] \text{for } ~ x & = \color{blue}{ 2.6444 } \Rightarrow p\left(2.6444\right) = \color{orangered}{ -4.6314 }\end{aligned} $$

So the turning points are:

$$ \begin{matrix} \left( 0.4561, -2.4187 \right) & \left( -0.6444, 2.6314 \right) & \left( 1.5439, 0.4187 \right) & \left( 2.6444, -4.6314 \right)\end{matrix} $$

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^3-60x^2+30x+10 $.

The zeros of second derivative are

$$ \begin{matrix}x_1 = 1 & x_2 = 2.2247 & x_3 = -0.2247 \end{matrix} $$

Substitute the x values into $ p(x) $ to get y coordinates

$$ \begin{aligned} \text{for } ~ x & = \color{blue}{ 1 } \Rightarrow p\left(1\right) = \color{orangered}{ -1 }\\[1 em] \text{for } ~ x & = \color{blue}{ 2.2247 } \Rightarrow p\left(2.2247\right) = \color{orangered}{ -2.5309 }\\[1 em] \text{for } ~ x & = \color{blue}{ -0.2247 } \Rightarrow p\left(-0.2247\right) = \color{orangered}{ 0.5309 }\end{aligned} $$

So the inflection points are:

$$ \begin{matrix} \left( 1, -1 \right) & \left( 2.2247, -2.5309 \right) & \left( -0.2247, 0.5309 \right)\end{matrix} $$

Graphing Polynomials