Sketch the graph of the polynomial function $$ p(t) = t^3-15t^2+72t $$

Tap the blue points to see coordinates.

STEP 1:Find the x-intercepts

To find the x-intercepts solve, the equation $ \color{blue}{ t^3-15t^2+72t = 0 } $

The solution of this equation is:

$$ \begin{matrix}t = 0 \end{matrix} $$

(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) = t^3-15t^2+72t } $, 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.

$$ \lim_{x \to -\infty} \left( t^3-15t^2+72t \right) = \lim_{x \to -\infty} t^3 = \color{blue}{ -\infty } $$

The graph starts in the lower-left corner.

$$ \lim_{x \to \infty} \left( t^3-15t^2+72t \right) = \lim_{x \to \infty} t^3 = \color{blue}{ \infty } $$

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) $:

$$ p^{\prime} (x) = 3t^2-30t+72 $$

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

$$ p^{\prime} (t) = 0 $$ $$ \begin{matrix}t_1 = 6 & t_2 = 4 \end{matrix} $$

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

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

$$ \begin{aligned} \text{for } ~ t & = \color{blue}{ 6 } \Rightarrow p\left(6\right) = \color{orangered}{ 108 }\\[1 em] \text{for } ~ t & = \color{blue}{ 4 } \Rightarrow p\left(4\right) = \color{orangered}{ 112 }\end{aligned} $$

So the turning points are:

$$ \begin{matrix} \left( 6, 108 \right) & \left( 4, 112 \right)\end{matrix} $$

STEP 5:Find the inflection points

The inflection points are located at zeroes of second derivative. The second derivative is $ p^{\prime \prime} (x) = 6t-30 $.

The zero of second derivative is

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

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

$$ \begin{aligned} \text{for } ~ t & = \color{blue}{ 5 } \Rightarrow p\left(5\right) = \color{orangered}{ 110 }\end{aligned} $$

So the inflection point is:

$$ \begin{matrix} \left( 5, 110 \right)\end{matrix} $$

Graphing Polynomials