Sketch the graph of the polynomial function $$ p(x) = x^6-100x^4+400x^2 $$

Tap the blue points to see coordinates.

STEP 1:Find the x-intercepts

To find the x-intercepts solve, the equation $ \color{blue}{ x^6-100x^4+400x^2 = 0 } $

The solutions of this equation are:

$$ \begin{matrix}x_1 = 0 & x_2 = 2.0431 & x_3 = -2.0431 & x_4 = -9.7891 & x_5 = 9.7891 \end{matrix} $$

(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^6-100x^4+400x^2 } $, 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.

$$ \lim_{x \to -\infty} \left( x^6-100x^4+400x^2 \right) = \lim_{x \to -\infty} x^6 = \color{blue}{ \infty } $$

The graph starts in the upper-left corner.

$$ \lim_{x \to \infty} \left( x^6-100x^4+400x^2 \right) = \lim_{x \to \infty} x^6 = \color{blue}{ \infty } $$

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) $:

$$ p^{\prime} (x) = 6x^5-400x^3+800x $$

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

$$ p^{\prime} (x) = 0 $$ $$ \begin{matrix}x_1 = 0 & x_2 = 1.4366 & x_3 = -1.4366 & x_4 = -8.0376 & x_5 = 8.0376 \end{matrix} $$

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

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

$$ \begin{aligned} \text{for } ~ x & = \color{blue}{ 0 } \Rightarrow p\left(0\right) = \color{orangered}{ 0 }\\[1 em] \text{for } ~ x & = \color{blue}{ 1.4366 } \Rightarrow p\left(1.4366\right) = \color{orangered}{ 408.3832 }\\[1 em] \text{for } ~ x & = \color{blue}{ -1.4366 } \Rightarrow p\left(-1.4366\right) = \color{orangered}{ 408.3832 }\\[1 em] \text{for } ~ x & = \color{blue}{ -8.0376 } \Rightarrow p\left(-8.0376\right) = \color{orangered}{ -121889.8647 }\\[1 em] \text{for } ~ x & = \color{blue}{ 8.0376 } \Rightarrow p\left(8.0376\right) = \color{orangered}{ -121889.8647 }\end{aligned} $$

So the turning points are:

$$ \begin{matrix} \left( 0, 0 \right) & \left( 1.4366, 408.3832 \right) & \left( -1.4366, 408.3832 \right) & \left( -8.0376, -121889.8647 \right) & \left( 8.0376, -121889.8647 \right)\end{matrix} $$

STEP 5:Find the inflection points

The inflection points are located at zeroes of second derivative. The second derivative is $ p^{\prime \prime} (x) = 30x^4-1200x^2+800 $.

The zeros of second derivative are

$$ \begin{matrix}x_1 = 0.8235 & x_2 = -0.8235 & x_3 = -6.2707 & x_4 = 6.2707 \end{matrix} $$

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

$$ \begin{aligned} \text{for } ~ x & = \color{blue}{ 0.8235 } \Rightarrow p\left(0.8235\right) = \color{orangered}{ 225.5881 }\\[1 em] \text{for } ~ x & = \color{blue}{ -0.8235 } \Rightarrow p\left(-0.8235\right) = \color{orangered}{ 225.5881 }\\[1 em] \text{for } ~ x & = \color{blue}{ -6.2707 } \Rightarrow p\left(-6.2707\right) = \color{orangered}{ -78092.1639 }\\[1 em] \text{for } ~ x & = \color{blue}{ 6.2707 } \Rightarrow p\left(6.2707\right) = \color{orangered}{ -78092.1639 }\end{aligned} $$

So the inflection points are:

$$ \begin{matrix} \left( 0.8235, 225.5881 \right) & \left( -0.8235, 225.5881 \right) & \left( -6.2707, -78092.1639 \right) & \left( 6.2707, -78092.1639 \right)\end{matrix} $$

Graphing Polynomials