Tap the blue points to see coordinates.
STEP 1:Find the x-intercepts
To find the x-intercepts solve, the equation $ \color{blue}{ 10t+ \cancel{5t} -\cancel{5t}+1-t^3 = 0 } $
The solutions of this equation are:
$$ \begin{matrix}t_1 = -0.1001 & t_2 = -3.111 & t_3 = 3.2111 \end{matrix} $$(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) = 10t+ \cancel{5t} -\cancel{5t}+1-t^3 } $, so:
$$ \text{Y inercept} = p(0) = 1 $$STEP 3:Find the end behavior
The end behavior of a polynomial is the same as the end behavior of a leading term.
$$ \lim_{x \to -\infty} \left( 10t+ \cancel{5t} -\cancel{5t}+1-t^3 \right) = \lim_{x \to -\infty} 10t = \color{blue}{ -\infty } $$The graph starts in the lower-left corner.
$$ \lim_{x \to \infty} \left( 10t+ \cancel{5t} -\cancel{5t}+1-t^3 \right) = \lim_{x \to \infty} 10t = \color{blue}{ \infty } $$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) $:
$$ p^{\prime} (x) = -3t^2+10 $$The x coordinate of the turning points are located at the zeros of the first derivative
$$ p^{\prime} (t) = 0 $$ $$ \begin{matrix}t_1 = \dfrac{\sqrt{ 30 }}{ 3 } & t_2 = - \dfrac{\sqrt{ 30 }}{ 3 } \end{matrix} $$(cleck here to see a explanation of how to solve the equation)
To find the y coordinates, substitute the above values into $ p(t) $
$$ \begin{aligned} \text{for } ~ t & = \color{blue}{ \frac{\sqrt{ 30 }}{ 3 } } \Rightarrow p\left(\frac{\sqrt{ 30 }}{ 3 }\right) = \color{orangered}{ 13.1716 }\\[1 em] \text{for } ~ t & = \color{blue}{ - \frac{\sqrt{ 30 }}{ 3 } } \Rightarrow p\left(- \frac{\sqrt{ 30 }}{ 3 }\right) = \color{orangered}{ -11.1716 }\end{aligned} $$So the turning points are:
$$ \begin{matrix} \left( \dfrac{\sqrt{ 30 }}{ 3 }, 13.1716 \right) & \left( - \dfrac{\sqrt{ 30 }}{ 3 }, -11.1716 \right)\end{matrix} $$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 $.
The zero of second derivative is
$$ \begin{matrix}t = 0 \end{matrix} $$Substitute the t value into $ p(t) $ to get y coordinates
$$ \begin{aligned} \text{for } ~ t & = \color{blue}{ 0 } \Rightarrow p\left(0\right) = \color{orangered}{ 1 }\end{aligned} $$So the inflection point is:
$$ \begin{matrix} \left( 0, 1 \right)\end{matrix} $$