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

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+2x^2+12 = 0 } $

Since above equation has no solutions we conclude that
polynomial has no x-intecepts.

(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+2x^2+12 } $, so:

$$ \text{Y inercept} = p(0) = 12 $$

STEP 3:Find the end behavior

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

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

The graph starts in the upper-left corner.

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

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) $:

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

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

$$ p^{\prime} (x) = 0 $$ $$ \begin{matrix}x_1 = 0 & x_2 = 1.6404 & x_3 = 0.6096 \end{matrix} $$

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

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

$$ \begin{aligned} \text{for } ~ x & = \color{blue}{ 0 } \Rightarrow p\left(0\right) = \color{orangered}{ 12 }\\[1 em] \text{for } ~ x & = \color{blue}{ 1.6404 } \Rightarrow p\left(1.6404\right) = \color{orangered}{ 11.3803 }\\[1 em] \text{for } ~ x & = \color{blue}{ 0.6096 } \Rightarrow p\left(0.6096\right) = \color{orangered}{ 12.2017 }\end{aligned} $$

So the turning points are:

$$ \begin{matrix} \left( 0, 12 \right) & \left( 1.6404, 11.3803 \right) & \left( 0.6096, 12.2017 \right)\end{matrix} $$

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+4 $.

The zeros of second derivative are

$$ \begin{matrix}x_1 = 1.2287 & x_2 = 0.2713 \end{matrix} $$

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

$$ \begin{aligned} \text{for } ~ x & = \color{blue}{ 1.2287 } \Rightarrow p\left(1.2287\right) = \color{orangered}{ 11.7337 }\\[1 em] \text{for } ~ x & = \color{blue}{ 0.2713 } \Rightarrow p\left(0.2713\right) = \color{orangered}{ 12.0927 }\end{aligned} $$

So the inflection points are:

$$ \begin{matrix} \left( 1.2287, 11.7337 \right) & \left( 0.2713, 12.0927 \right)\end{matrix} $$

Graphing Polynomials