Tap the blue points to see coordinates.
STEP 1:Find the x-intercepts
To find the x-intercepts solve, the equation $ \color{blue}{ -\frac{1}{10}x^4+\frac{2}{10}x^3+\frac{13}{10}x^2-\frac{14}{10}x-\frac{24}{10} = 0 } $
The solutions of this equation are:
$$ \begin{matrix}x_1 = 2 & x_2 = 4 & x_3 = -1 & x_4 = -3 \end{matrix} $$(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) = -\frac{1}{10}x^4+\frac{2}{10}x^3+\frac{13}{10}x^2-\frac{14}{10}x-\frac{24}{10} } $, so:
$$ \text{Y inercept} = p(0) = -\frac{ 12 }{ 5 } $$STEP 3:Find the end behavior
The end behavior of a polynomial is the same as the end behavior of a leading term.
$$ \lim_{x \to -\infty} \left( -\frac{1}{10}x^4+\frac{2}{10}x^3+\frac{13}{10}x^2-\frac{14}{10}x-\frac{24}{10} \right) = \lim_{x \to -\infty} -\frac{1}{10}x^4 = \color{blue}{ -\infty } $$The graph starts in the lower-left corner.
$$ \lim_{x \to \infty} \left( -\frac{1}{10}x^4+\frac{2}{10}x^3+\frac{13}{10}x^2-\frac{14}{10}x-\frac{24}{10} \right) = \lim_{x \to \infty} -\frac{1}{10}x^4 = \color{blue}{ -\infty } $$The graph ends in the lower-right corner.
STEP 4:Find the turning points
To determine the turning points, we need to find the first derivative of $ p(x) $:
$$ p^{\prime} (x) = -\frac{2}{5}x^3+\frac{3}{5}x^2+\frac{13}{5}x-\frac{7}{5} $$The x coordinate of the turning points are located at the zeros of the first derivative
$$ p^{\prime} (x) = 0 $$ $$ \begin{matrix}x_1 = \dfrac{ 1 }{ 2 } & x_2 = 3.1926 & x_3 = -2.1926 \end{matrix} $$(cleck here to see a explanation of how to solve the equation)
To find the y coordinates, substitute the above values into $ p(x) $
$$ \begin{aligned} \text{for } ~ x & = \color{blue}{ \frac{ 1 }{ 2 } } \Rightarrow p\left(\frac{ 1 }{ 2 }\right) = \color{orangered}{ -\frac{ 441 }{ 160 } }\\[1 em] \text{for } ~ x & = \color{blue}{ 3.1926 } \Rightarrow p\left(3.1926\right) = \color{orangered}{ 2.5 }\\[1 em] \text{for } ~ x & = \color{blue}{ -2.1926 } \Rightarrow p\left(-2.1926\right) = \color{orangered}{ 2.5 }\end{aligned} $$So the turning points are:
$$ \begin{matrix} \left( \dfrac{ 1 }{ 2 }, -\dfrac{ 441 }{ 160 } \right) & \left( 3.1926, 2.5 \right) & \left( -2.1926, 2.5 \right)\end{matrix} $$STEP 5:Find the inflection points
The inflection points are located at zeroes of second derivative. The second derivative is $ p^{\prime \prime} (x) = -\frac{6}{5}x^2+\frac{6}{5}x+\frac{13}{5} $.
The zeros of second derivative are
$$ \begin{matrix}x_1 = 2.0546 & x_2 = -1.0546 \end{matrix} $$Substitute the x values into $ p(x) $ to get y coordinates
$$ \begin{aligned} \text{for } ~ x & = \color{blue}{ 2.0546 } \Rightarrow p\left(2.0546\right) = \color{orangered}{ 0.1639 }\\[1 em] \text{for } ~ x & = \color{blue}{ -1.0546 } \Rightarrow p\left(-1.0546\right) = \color{orangered}{ 0.1639 }\end{aligned} $$So the inflection points are:
$$ \begin{matrix} \left( 2.0546, 0.1639 \right) & \left( -1.0546, 0.1639 \right)\end{matrix} $$