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-x^3-13x^2+x+12 = 0 } $
The solutions of this equation are:
$$ \begin{matrix}x_1 = 1 & x_2 = 4 & x_3 = -1 & x_4 = -3 \end{matrix} $$(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) = x^4-x^3-13x^2+x+12 } $, so:
$$ \text{Y inercept} = p(0) = 12 $$STEP 3:Find the end behavior
The end behavior of a polynomial is the same as the end behavior of a leading term.
$$ \lim_{x \to -\infty} \left( x^4-x^3-13x^2+x+12 \right) = \lim_{x \to -\infty} x^4 = \color{blue}{ \infty } $$The graph starts in the upper-left corner.
$$ \lim_{x \to \infty} \left( x^4-x^3-13x^2+x+12 \right) = \lim_{x \to \infty} x^4 = \color{blue}{ \infty } $$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) $:
$$ p^{\prime} (x) = 4x^3-3x^2-26x+1 $$The x coordinate of the turning points are located at the zeros of the first derivative
$$ p^{\prime} (x) = 0 $$ $$ \begin{matrix}x_1 = 0.0383 & x_2 = -2.2237 & x_3 = 2.9354 \end{matrix} $$(cleck here to see a explanation of how to solve the equation)
To find the y coordinates, substitute the above values into $ p(x) $
$$ \begin{aligned} \text{for } ~ x & = \color{blue}{ 0.0383 } \Rightarrow p\left(0.0383\right) = \color{orangered}{ 12.0192 }\\[1 em] \text{for } ~ x & = \color{blue}{ -2.2237 } \Rightarrow p\left(-2.2237\right) = \color{orangered}{ -19.0593 }\\[1 em] \text{for } ~ x & = \color{blue}{ 2.9354 } \Rightarrow p\left(2.9354\right) = \color{orangered}{ -48.1278 }\end{aligned} $$So the turning points are:
$$ \begin{matrix} \left( 0.0383, 12.0192 \right) & \left( -2.2237, -19.0593 \right) & \left( 2.9354, -48.1278 \right)\end{matrix} $$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^2-6x-26 $.
The zeros of second derivative are
$$ \begin{matrix}x_1 = 1.743 & x_2 = -1.243 \end{matrix} $$Substitute the x values into $ p(x) $ to get y coordinates
$$ \begin{aligned} \text{for } ~ x & = \color{blue}{ 1.743 } \Rightarrow p\left(1.743\right) = \color{orangered}{ -21.8185 }\\[1 em] \text{for } ~ x & = \color{blue}{ -1.243 } \Rightarrow p\left(-1.243\right) = \color{orangered}{ -5.0218 }\end{aligned} $$So the inflection points are:
$$ \begin{matrix} \left( 1.743, -21.8185 \right) & \left( -1.243, -5.0218 \right)\end{matrix} $$