Tap the blue points to see coordinates.
STEP 1:Find the x-intercepts
To find the x-intercepts solve, the equation $ \color{blue}{ 4x^4-1+x^5-3x^2 = 0 } $
The solutions of this equation are:
$$ \begin{matrix}x_1 = -1.148 & x_2 = 0.9214 & x_3 = -3.7858 \end{matrix} $$(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) = 4x^4-1+x^5-3x^2 } $, 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.
$$ \lim_{x \to -\infty} \left( 4x^4-1+x^5-3x^2 \right) = \lim_{x \to -\infty} 4x^4 = \color{blue}{ \infty } $$The graph starts in the upper-left corner.
$$ \lim_{x \to \infty} \left( 4x^4-1+x^5-3x^2 \right) = \lim_{x \to \infty} 4x^4 = \color{blue}{ \infty } $$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) $:
$$ p^{\prime} (x) = 5x^4+16x^3-6x $$The x coordinate of the turning points are located at the zeros of the first derivative
$$ p^{\prime} (x) = 0 $$ $$ \begin{matrix}x_1 = 0 & x_2 = 0.5646 & x_3 = -0.6917 & x_4 = -3.0729 \end{matrix} $$(cleck here to see a explanation of how to solve the equation)
To find the y coordinates, substitute the above values into $ p(x) $
$$ \begin{aligned} \text{for } ~ x & = \color{blue}{ 0 } \Rightarrow p\left(0\right) = \color{orangered}{ -1 }\\[1 em] \text{for } ~ x & = \color{blue}{ 0.5646 } \Rightarrow p\left(0.5646\right) = \color{orangered}{ -1.4925 }\\[1 em] \text{for } ~ x & = \color{blue}{ -0.6917 } \Rightarrow p\left(-0.6917\right) = \color{orangered}{ -1.678 }\\[1 em] \text{for } ~ x & = \color{blue}{ -3.0729 } \Rightarrow p\left(-3.0729\right) = \color{orangered}{ 53.3366 }\end{aligned} $$So the turning points are:
$$ \begin{matrix} \left( 0, -1 \right) & \left( 0.5646, -1.4925 \right) & \left( -0.6917, -1.678 \right) & \left( -3.0729, 53.3366 \right)\end{matrix} $$STEP 5:Find the inflection points
The inflection points are located at zeroes of second derivative. The second derivative is $ p^{\prime \prime} (x) = 20x^3+48x^2-6 $.
The zeros of second derivative are
$$ \begin{matrix}x_1 = 0.3314 & x_2 = -0.3859 & x_3 = -2.3455 \end{matrix} $$Substitute the x values into $ p(x) $ to get y coordinates
$$ \begin{aligned} \text{for } ~ x & = \color{blue}{ 0.3314 } \Rightarrow p\left(0.3314\right) = \color{orangered}{ -1.2772 }\\[1 em] \text{for } ~ x & = \color{blue}{ -0.3859 } \Rightarrow p\left(-0.3859\right) = \color{orangered}{ -1.3667 }\\[1 em] \text{for } ~ x & = \color{blue}{ -2.3455 } \Rightarrow p\left(-2.3455\right) = \color{orangered}{ 32.5682 }\end{aligned} $$So the inflection points are:
$$ \begin{matrix} \left( 0.3314, -1.2772 \right) & \left( -0.3859, -1.3667 \right) & \left( -2.3455, 32.5682 \right)\end{matrix} $$