Sketch the graph of the polynomial function $$ p(x) = 3x^4-7x^3-6x^2+12x+8 $$

Tap the blue points to see coordinates.

STEP 1:Find the x-intercepts

To find the x-intercepts solve, the equation $ \color{blue}{ 3x^4-7x^3-6x^2+12x+8 = 0 } $

The solutions of this equation are:

$$ \begin{matrix}x_1 = 2 & x_2 = -1 & x_3 = -\dfrac{ 2 }{ 3 } & x_4 = 2 \end{matrix} $$

(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 $ x = 0 $ into $ \color{blue}{ p(x) = 3x^4-7x^3-6x^2+12x+8 } $, so:

$$ \text{Y inercept} = p(0) = 8 $$

STEP 3:Find the end behavior

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

$$ \lim_{x \to -\infty} \left( 3x^4-7x^3-6x^2+12x+8 \right) = \lim_{x \to -\infty} 3x^4 = \color{blue}{ \infty } $$

The graph starts in the upper-left corner.

$$ \lim_{x \to \infty} \left( 3x^4-7x^3-6x^2+12x+8 \right) = \lim_{x \to \infty} 3x^4 = \color{blue}{ \infty } $$

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) $:

$$ p^{\prime} (x) = 12x^3-21x^2-12x+12 $$

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

$$ p^{\prime} (x) = 0 $$ $$ \begin{matrix}x_1 = 2 & x_2 = 0.5931 & x_3 = -0.8431 \end{matrix} $$

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

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

$$ \begin{aligned} \text{for } ~ x & = \color{blue}{ 2 } \Rightarrow p\left(2\right) = \color{orangered}{ 0 }\\[1 em] \text{for } ~ x & = \color{blue}{ 0.5931 } \Rightarrow p\left(0.5931\right) = \color{orangered}{ 11.9174 }\\[1 em] \text{for } ~ x & = \color{blue}{ -0.8431 } \Rightarrow p\left(-0.8431\right) = \color{orangered}{ -0.6713 }\end{aligned} $$

So the turning points are:

$$ \begin{matrix} \left( 2, 0 \right) & \left( 0.5931, 11.9174 \right) & \left( -0.8431, -0.6713 \right)\end{matrix} $$

STEP 5:Find the inflection points

The inflection points are located at zeroes of second derivative. The second derivative is $ p^{\prime \prime} (x) = 36x^2-42x-12 $.

The zeros of second derivative are

$$ \begin{matrix}x_1 = 1.4041 & x_2 = -0.2374 \end{matrix} $$

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

$$ \begin{aligned} \text{for } ~ x & = \color{blue}{ 1.4041 } \Rightarrow p\left(1.4041\right) = \color{orangered}{ 5.3037 }\\[1 em] \text{for } ~ x & = \color{blue}{ -0.2374 } \Rightarrow p\left(-0.2374\right) = \color{orangered}{ 4.9162 }\end{aligned} $$

So the inflection points are:

$$ \begin{matrix} \left( 1.4041, 5.3037 \right) & \left( -0.2374, 4.9162 \right)\end{matrix} $$

Graphing Polynomials