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

Tap the blue points to see coordinates.

STEP 1:Find the x-intercepts

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

The solutions of this equation are:

$$ \begin{matrix}n_1 = 0 & n_2 = -\dfrac{ 1 }{ 2 } \end{matrix} $$

(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 $ n = 0 $ into $ \color{blue}{ p(n) = 2n^4+n^3 } $, 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.

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

The graph starts in the upper-left corner.

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

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(n) $:

$$ p^{\prime} (x) = 8n^3+3n^2 $$

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

$$ p^{\prime} (n) = 0 $$ $$ \begin{matrix}n_1 = 0 & n_2 = -\dfrac{ 3 }{ 8 } \end{matrix} $$

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

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

$$ \begin{aligned} \text{for } ~ n & = \color{blue}{ 0 } \Rightarrow p\left(0\right) = \color{orangered}{ 0 }\\[1 em] \text{for } ~ n & = \color{blue}{ -\frac{ 3 }{ 8 } } \Rightarrow p\left(-\frac{ 3 }{ 8 }\right) = \color{orangered}{ -\frac{ 27 }{ 2048 } }\end{aligned} $$

So the turning points are:

$$ \begin{matrix} \left( 0, 0 \right) & \left( -\dfrac{ 3 }{ 8 }, -\dfrac{ 27 }{ 2048 } \right)\end{matrix} $$

STEP 5:Find the inflection points

The inflection points are located at zeroes of second derivative. The second derivative is $ p^{\prime \prime} (x) = 24n^2+6n $.

The zeros of second derivative are

$$ \begin{matrix}n_1 = 0 & n_2 = -\dfrac{ 1 }{ 4 } \end{matrix} $$

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

$$ \begin{aligned} \text{for } ~ n & = \color{blue}{ 0 } \Rightarrow p\left(0\right) = \color{orangered}{ 0 }\\[1 em] \text{for } ~ n & = \color{blue}{ -\frac{ 1 }{ 4 } } \Rightarrow p\left(-\frac{ 1 }{ 4 }\right) = \color{orangered}{ -\frac{ 1 }{ 128 } }\end{aligned} $$

So the inflection points are:

$$ \begin{matrix} \left( 0, 0 \right) & \left( -\dfrac{ 1 }{ 4 }, -\dfrac{ 1 }{ 128 } \right)\end{matrix} $$

Graphing Polynomials