Sketch the graph of the polynomial function $$ p(x) = x^4-12x^3+46x^2-60x+25 $$

Tap the blue points to see coordinates.

STEP 1:Find the x-intercepts

To find the x-intercepts solve, the equation $ \color{blue}{ x^4-12x^3+46x^2-60x+25 = 0 } $

The solutions of this equation are:

$$ \begin{matrix}x_1 = 1 & x_2 = 5 & x_3 = 1 & x_4 = 5 \end{matrix} $$

(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^4-12x^3+46x^2-60x+25 } $, so:

$$ \text{Y inercept} = p(0) = 25 $$

STEP 3:Find the end behavior

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

$$ \lim_{x \to -\infty} \left( x^4-12x^3+46x^2-60x+25 \right) = \lim_{x \to -\infty} x^4 = \color{blue}{ \infty } $$

The graph starts in the upper-left corner.

$$ \lim_{x \to \infty} \left( x^4-12x^3+46x^2-60x+25 \right) = \lim_{x \to \infty} x^4 = \color{blue}{ \infty } $$

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) $:

$$ p^{\prime} (x) = 4x^3-36x^2+92x-60 $$

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

$$ p^{\prime} (x) = 0 $$ $$ \begin{matrix}x_1 = 1 & x_2 = 3 & x_3 = 5 \end{matrix} $$

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

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

$$ \begin{aligned} \text{for } ~ x & = \color{blue}{ 1 } \Rightarrow p\left(1\right) = \color{orangered}{ 0 }\\[1 em] \text{for } ~ x & = \color{blue}{ 3 } \Rightarrow p\left(3\right) = \color{orangered}{ 16 }\\[1 em] \text{for } ~ x & = \color{blue}{ 5 } \Rightarrow p\left(5\right) = \color{orangered}{ 0 }\end{aligned} $$

So the turning points are:

$$ \begin{matrix} \left( 1, 0 \right) & \left( 3, 16 \right) & \left( 5, 0 \right)\end{matrix} $$

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^2-72x+92 $.

The zeros of second derivative are

$$ \begin{matrix}x_1 = 4.1547 & x_2 = 1.8453 \end{matrix} $$

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

$$ \begin{aligned} \text{for } ~ x & = \color{blue}{ 4.1547 } \Rightarrow p\left(4.1547\right) = \color{orangered}{ 7.1111 }\\[1 em] \text{for } ~ x & = \color{blue}{ 1.8453 } \Rightarrow p\left(1.8453\right) = \color{orangered}{ 7.1111 }\end{aligned} $$

So the inflection points are:

$$ \begin{matrix} \left( 4.1547, 7.1111 \right) & \left( 1.8453, 7.1111 \right)\end{matrix} $$

Graphing Polynomials