Tap the blue points to see coordinates.
STEP 1:Find the x-intercepts
To find the x-intercepts solve, the equation $ \color{blue}{ -x^4-3x^3+4x^2+12x = 0 } $
The solutions of this equation are:
$$ \begin{matrix}x_1 = 0 & x_2 = 2 & x_3 = -2 & x_4 = -3 \end{matrix} $$(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) = -x^4-3x^3+4x^2+12x } $, 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.
$$ \lim_{x \to -\infty} \left( -x^4-3x^3+4x^2+12x \right) = \lim_{x \to -\infty} -x^4 = \color{blue}{ -\infty } $$The graph starts in the lower-left corner.
$$ \lim_{x \to \infty} \left( -x^4-3x^3+4x^2+12x \right) = \lim_{x \to \infty} -x^4 = \color{blue}{ -\infty } $$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) $:
$$ p^{\prime} (x) = -4x^3-9x^2+8x+12 $$The x coordinate of the turning points are located at the zeros of the first derivative
$$ p^{\prime} (x) = 0 $$ $$ \begin{matrix}x_1 = 1.2538 & x_2 = -0.9295 & x_3 = -2.5742 \end{matrix} $$(cleck here to see a explanation of how to solve the equation)
To find the y coordinates, substitute the above values into $ p(x) $
$$ \begin{aligned} \text{for } ~ x & = \color{blue}{ 1.2538 } \Rightarrow p\left(1.2538\right) = \color{orangered}{ 12.9495 }\\[1 em] \text{for } ~ x & = \color{blue}{ -0.9295 } \Rightarrow p\left(-0.9295\right) = \color{orangered}{ -6.0354 }\\[1 em] \text{for } ~ x & = \color{blue}{ -2.5742 } \Rightarrow p\left(-2.5742\right) = \color{orangered}{ 2.8789 }\end{aligned} $$So the turning points are:
$$ \begin{matrix} \left( 1.2538, 12.9495 \right) & \left( -0.9295, -6.0354 \right) & \left( -2.5742, 2.8789 \right)\end{matrix} $$STEP 5:Find the inflection points
The inflection points are located at zeroes of second derivative. The second derivative is $ p^{\prime \prime} (x) = -12x^2-18x+8 $.
The zeros of second derivative are
$$ \begin{matrix}x_1 = 0.3587 & x_2 = -1.8587 \end{matrix} $$Substitute the x values into $ p(x) $ to get y coordinates
$$ \begin{aligned} \text{for } ~ x & = \color{blue}{ 0.3587 } \Rightarrow p\left(0.3587\right) = \color{orangered}{ 4.6638 }\\[1 em] \text{for } ~ x & = \color{blue}{ -1.8587 } \Rightarrow p\left(-1.8587\right) = \color{orangered}{ -1.1568 }\end{aligned} $$So the inflection points are:
$$ \begin{matrix} \left( 0.3587, 4.6638 \right) & \left( -1.8587, -1.1568 \right)\end{matrix} $$