Sketch the graph of the polynomial function $$ p(x) = x^4-6x^3+8x^2+2x-1 $$

Tap the blue points to see coordinates.

STEP 1:Find the x-intercepts

To find the x-intercepts solve, the equation $ \color{blue}{ x^4-6x^3+8x^2+2x-1 = 0 } $

The solutions of this equation are:

$$ \begin{matrix}x_1 = 0.268 & x_2 = -0.4142 & x_3 = 2.4142 & x_4 = 3.7321 \end{matrix} $$

(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^4-6x^3+8x^2+2x-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.

$$ \lim_{x \to -\infty} \left( x^4-6x^3+8x^2+2x-1 \right) = \lim_{x \to -\infty} x^4 = \color{blue}{ \infty } $$

The graph starts in the upper-left corner.

$$ \lim_{x \to \infty} \left( x^4-6x^3+8x^2+2x-1 \right) = \lim_{x \to \infty} x^4 = \color{blue}{ \infty } $$

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) $:

$$ p^{\prime} (x) = 4x^3-18x^2+16x+2 $$

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

$$ p^{\prime} (x) = 0 $$ $$ \begin{matrix}x_1 = -0.1108 & x_2 = 1.4088 & x_3 = 3.202 \end{matrix} $$

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

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

$$ \begin{aligned} \text{for } ~ x & = \color{blue}{ -0.1108 } \Rightarrow p\left(-0.1108\right) = \color{orangered}{ -1.1151 }\\[1 em] \text{for } ~ x & = \color{blue}{ 1.4088 } \Rightarrow p\left(1.4088\right) = \color{orangered}{ 4.858 }\\[1 em] \text{for } ~ x & = \color{blue}{ 3.202 } \Rightarrow p\left(3.202\right) = \color{orangered}{ -4.4304 }\end{aligned} $$

So the turning points are:

$$ \begin{matrix} \left( -0.1108, -1.1151 \right) & \left( 1.4088, 4.858 \right) & \left( 3.202, -4.4304 \right)\end{matrix} $$

STEP 5:Find the inflection points

The inflection points are located at zeroes of second derivative. The second derivative is $ p^{\prime \prime} (x) = 12x^2-36x+16 $.

The zeros of second derivative are

$$ \begin{matrix}x_1 = 2.4574 & x_2 = 0.5426 \end{matrix} $$

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

$$ \begin{aligned} \text{for } ~ x & = \color{blue}{ 2.4574 } \Rightarrow p\left(2.4574\right) = \color{orangered}{ -0.3463 }\\[1 em] \text{for } ~ x & = \color{blue}{ 0.5426 } \Rightarrow p\left(0.5426\right) = \color{orangered}{ 1.5685 }\end{aligned} $$

So the inflection points are:

$$ \begin{matrix} \left( 2.4574, -0.3463 \right) & \left( 0.5426, 1.5685 \right)\end{matrix} $$

Graphing Polynomials