Sketch the graph of the polynomial function $$ p(t) = 4t^3-5t^2-6t $$

Tap the blue points to see coordinates.

STEP 1:Find the x-intercepts

To find the x-intercepts solve, the equation $ \color{blue}{ 4t^3-5t^2-6t = 0 } $

The solutions of this equation are:

$$ \begin{matrix}t_1 = 0 & t_2 = 2 & t_3 = -\dfrac{ 3 }{ 4 } \end{matrix} $$

(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 $ t = 0 $ into $ \color{blue}{ p(t) = 4t^3-5t^2-6t } $, 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.

$$ \lim_{x \to -\infty} \left( 4t^3-5t^2-6t \right) = \lim_{x \to -\infty} 4t^3 = \color{blue}{ -\infty } $$

The graph starts in the lower-left corner.

$$ \lim_{x \to \infty} \left( 4t^3-5t^2-6t \right) = \lim_{x \to \infty} 4t^3 = \color{blue}{ \infty } $$

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(t) $:

$$ p^{\prime} (x) = 12t^2-10t-6 $$

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

$$ p^{\prime} (t) = 0 $$ $$ \begin{matrix}t_1 = 1.2374 & t_2 = -0.4041 \end{matrix} $$

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

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

$$ \begin{aligned} \text{for } ~ t & = \color{blue}{ 1.2374 } \Rightarrow p\left(1.2374\right) = \color{orangered}{ -7.5016 }\\[1 em] \text{for } ~ t & = \color{blue}{ -0.4041 } \Rightarrow p\left(-0.4041\right) = \color{orangered}{ 1.3442 }\end{aligned} $$

So the turning points are:

$$ \begin{matrix} \left( 1.2374, -7.5016 \right) & \left( -0.4041, 1.3442 \right)\end{matrix} $$

STEP 5:Find the inflection points

The inflection points are located at zeroes of second derivative. The second derivative is $ p^{\prime \prime} (x) = 24t-10 $.

The zero of second derivative is

$$ \begin{matrix}t = \dfrac{ 5 }{ 12 } \end{matrix} $$

Substitute the t value into $ p(t) $ to get y coordinates

$$ \begin{aligned} \text{for } ~ t & = \color{blue}{ \frac{ 5 }{ 12 } } \Rightarrow p\left(\frac{ 5 }{ 12 }\right) = \color{orangered}{ -\frac{ 665 }{ 216 } }\end{aligned} $$

So the inflection point is:

$$ \begin{matrix} \left( \dfrac{ 5 }{ 12 }, -\dfrac{ 665 }{ 216 } \right)\end{matrix} $$

Graphing Polynomials