Sketch the graph of the polynomial function $$ p(x) = 3x^4-2x^3-16x^2-14x-3 $$

Tap the blue points to see coordinates.

STEP 1:Find the x-intercepts

To find the x-intercepts solve, the equation $ \color{blue}{ 3x^4-2x^3-16x^2-14x-3 = 0 } $

The solutions of this equation are:

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

(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^4-2x^3-16x^2-14x-3 } $, so:

$$ \text{Y inercept} = p(0) = -3 $$

STEP 3:Find the end behavior

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

$$ \lim_{x \to -\infty} \left( 3x^4-2x^3-16x^2-14x-3 \right) = \lim_{x \to -\infty} 3x^4 = \color{blue}{ \infty } $$

The graph starts in the upper-left corner.

$$ \lim_{x \to \infty} \left( 3x^4-2x^3-16x^2-14x-3 \right) = \lim_{x \to \infty} 3x^4 = \color{blue}{ \infty } $$

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) $:

$$ p^{\prime} (x) = 12x^3-6x^2-32x-14 $$

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

$$ p^{\prime} (x) = 0 $$ $$ \begin{matrix}x_1 = -1 & x_2 = 2.065 & x_3 = -0.565 \end{matrix} $$

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

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

$$ \begin{aligned} \text{for } ~ x & = \color{blue}{ -1 } \Rightarrow p\left(-1\right) = \color{orangered}{ 0 }\\[1 em] \text{for } ~ x & = \color{blue}{ 2.065 } \Rightarrow p\left(2.065\right) = \color{orangered}{ -63.198 }\\[1 em] \text{for } ~ x & = \color{blue}{ -0.565 } \Rightarrow p\left(-0.565\right) = \color{orangered}{ 0.4688 }\end{aligned} $$

So the turning points are:

$$ \begin{matrix} \left( -1, 0 \right) & \left( 2.065, -63.198 \right) & \left( -0.565, 0.4688 \right)\end{matrix} $$

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^2-12x-32 $.

The zeros of second derivative are

$$ \begin{matrix}x_1 = 1.1241 & x_2 = -0.7908 \end{matrix} $$

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

$$ \begin{aligned} \text{for } ~ x & = \color{blue}{ 1.1241 } \Rightarrow p\left(1.1241\right) = \color{orangered}{ -37.0055 }\\[1 em] \text{for } ~ x & = \color{blue}{ -0.7908 } \Rightarrow p\left(-0.7908\right) = \color{orangered}{ 0.2277 }\end{aligned} $$

So the inflection points are:

$$ \begin{matrix} \left( 1.1241, -37.0055 \right) & \left( -0.7908, 0.2277 \right)\end{matrix} $$

Graphing Polynomials