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

Tap the blue points to see coordinates.

STEP 1:Find the x-intercepts

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

The solutions of this equation are:

$$ \begin{matrix}x_1 = 0 & x_2 = 1 & x_3 = 1 & x_4 = 1 \end{matrix} $$

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

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

STEP 3:Find the end behavior

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

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

The graph starts in the upper-left corner.

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

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) $:

$$ p^{\prime} (x) = 4x^3-9x^2+6x-1 $$

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

$$ p^{\prime} (x) = 0 $$ $$ \begin{matrix}x_1 = 1 & x_2 = \dfrac{ 1 }{ 4 } & x_3 = 1 \end{matrix} $$

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

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

$$ \begin{aligned} \text{for } ~ x & = \color{blue}{ 1 } \Rightarrow p\left(1\right) = \color{orangered}{ 0 }\\[1 em] \text{for } ~ x & = \color{blue}{ \frac{ 1 }{ 4 } } \Rightarrow p\left(\frac{ 1 }{ 4 }\right) = \color{orangered}{ -\frac{ 27 }{ 256 } }\\[1 em] \text{for } ~ x & = \color{blue}{ 1 } \Rightarrow p\left(1\right) = \color{orangered}{ 0 }\end{aligned} $$

So the turning points are:

$$ \begin{matrix} \left( 1, 0 \right) & \left( \dfrac{ 1 }{ 4 }, -\dfrac{ 27 }{ 256 } \right) & \left( 1, 0 \right)\end{matrix} $$

STEP 5:Find the inflection points

The inflection points are located at zeroes of second derivative. The second derivative is $ p^{\prime \prime} (x) = 12x^2-18x+6 $.

The zeros of second derivative are

$$ \begin{matrix}x_1 = 1 & x_2 = \dfrac{ 1 }{ 2 } \end{matrix} $$

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

$$ \begin{aligned} \text{for } ~ x & = \color{blue}{ 1 } \Rightarrow p\left(1\right) = \color{orangered}{ 0 }\\[1 em] \text{for } ~ x & = \color{blue}{ \frac{ 1 }{ 2 } } \Rightarrow p\left(\frac{ 1 }{ 2 }\right) = \color{orangered}{ -\frac{ 1 }{ 16 } }\end{aligned} $$

So the inflection points are:

$$ \begin{matrix} \left( 1, 0 \right) & \left( \dfrac{ 1 }{ 2 }, -\dfrac{ 1 }{ 16 } \right)\end{matrix} $$

Graphing Polynomials