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

Tap the blue points to see coordinates.

STEP 1:Find the x-intercepts

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

The solutions of this equation are:

$$ \begin{matrix}x_1 = 1 & x_2 = 2 & x_3 = -1 & x_4 = 2 \end{matrix} $$

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

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

The graph starts in the upper-left corner.

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

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) $:

$$ p^{\prime} (x) = 4x^3-12x^2+6x+4 $$

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

$$ p^{\prime} (x) = 0 $$ $$ \begin{matrix}x_1 = 2 & x_2 = 1.366 & x_3 = -0.366 \end{matrix} $$

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

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

$$ \begin{aligned} \text{for } ~ x & = \color{blue}{ 2 } \Rightarrow p\left(2\right) = \color{orangered}{ 0 }\\[1 em] \text{for } ~ x & = \color{blue}{ 1.366 } \Rightarrow p\left(1.366\right) = \color{orangered}{ 0.3481 }\\[1 em] \text{for } ~ x & = \color{blue}{ -0.366 } \Rightarrow p\left(-0.366\right) = \color{orangered}{ -4.8481 }\end{aligned} $$

So the turning points are:

$$ \begin{matrix} \left( 2, 0 \right) & \left( 1.366, 0.3481 \right) & \left( -0.366, -4.8481 \right)\end{matrix} $$

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^2-24x+6 $.

The zeros of second derivative are

$$ \begin{matrix}x_1 = 1.7071 & x_2 = 0.2929 \end{matrix} $$

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

$$ \begin{aligned} \text{for } ~ x & = \color{blue}{ 1.7071 } \Rightarrow p\left(1.7071\right) = \color{orangered}{ 0.1642 }\\[1 em] \text{for } ~ x & = \color{blue}{ 0.2929 } \Rightarrow p\left(0.2929\right) = \color{orangered}{ -2.6642 }\end{aligned} $$

So the inflection points are:

$$ \begin{matrix} \left( 1.7071, 0.1642 \right) & \left( 0.2929, -2.6642 \right)\end{matrix} $$

Graphing Polynomials