Tap the blue points to see coordinates.
STEP 1:Find the x-intercepts
To find the x-intercepts solve, the equation $ \color{blue}{ 2x^6+6x^5-12x^3-6x^2+6x+4 = 0 } $
The solutions of this equation are:
$$ \begin{matrix}x_1 = 1 & x_2 = -1 & x_3 = -2 & x_4 = 1 & x_5 = -1 & x_6 = -1 \end{matrix} $$(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) = 2x^6+6x^5-12x^3-6x^2+6x+4 } $, so:
$$ \text{Y inercept} = p(0) = 4 $$STEP 3:Find the end behavior
The end behavior of a polynomial is the same as the end behavior of a leading term.
$$ \lim_{x \to -\infty} \left( 2x^6+6x^5-12x^3-6x^2+6x+4 \right) = \lim_{x \to -\infty} 2x^6 = \color{blue}{ \infty } $$The graph starts in the upper-left corner.
$$ \lim_{x \to \infty} \left( 2x^6+6x^5-12x^3-6x^2+6x+4 \right) = \lim_{x \to \infty} 2x^6 = \color{blue}{ \infty } $$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) $:
$$ p^{\prime} (x) = 12x^5+30x^4-36x^2-12x+6 $$The x coordinate of the turning points are located at the zeros of the first derivative
$$ p^{\prime} (x) = 0 $$ $$ \begin{matrix}x_1 = 1 & x_2 = -1 & x_3 = -1 & x_4 = 0.2808 & x_5 = -1.7808 \end{matrix} $$(cleck here to see a explanation of how to solve the equation)
To find the y coordinates, substitute the above values into $ p(x) $
$$ \begin{aligned} \text{for } ~ x & = \color{blue}{ 1 } \Rightarrow p\left(1\right) = \color{orangered}{ 0 }\\[1 em] \text{for } ~ x & = \color{blue}{ -1 } \Rightarrow p\left(-1\right) = \color{orangered}{ 0 }\\[1 em] \text{for } ~ x & = \color{blue}{ -1 } \Rightarrow p\left(-1\right) = \color{orangered}{ 0 }\\[1 em] \text{for } ~ x & = \color{blue}{ 0.2808 } \Rightarrow p\left(0.2808\right) = \color{orangered}{ 4.9575 }\\[1 em] \text{for } ~ x & = \color{blue}{ -1.7808 } \Rightarrow p\left(-1.7808\right) = \color{orangered}{ -1.6137 }\end{aligned} $$So the turning points are:
$$ \begin{matrix} \left( 1, 0 \right) & \left( -1, 0 \right) & \left( -1, 0 \right) & \left( 0.2808, 4.9575 \right) & \left( -1.7808, -1.6137 \right)\end{matrix} $$STEP 5:Find the inflection points
The inflection points are located at zeroes of second derivative. The second derivative is $ p^{\prime \prime} (x) = 60x^4+120x^3-72x-12 $.
The zeros of second derivative are
$$ \begin{matrix}x_1 = -1 & x_2 = -0.1748 & x_3 = 0.7339 & x_4 = -1.5591 \end{matrix} $$Substitute the x values into $ p(x) $ to get y coordinates
$$ \begin{aligned} \text{for } ~ x & = \color{blue}{ -1 } \Rightarrow p\left(-1\right) = \color{orangered}{ 0 }\\[1 em] \text{for } ~ x & = \color{blue}{ -0.1748 } \Rightarrow p\left(-0.1748\right) = \color{orangered}{ 2.8311 }\\[1 em] \text{for } ~ x & = \color{blue}{ 0.7339 } \Rightarrow p\left(0.7339\right) = \color{orangered}{ 2.0182 }\\[1 em] \text{for } ~ x & = \color{blue}{ -1.5591 } \Rightarrow p\left(-1.5591\right) = \color{orangered}{ -1.0093 }\end{aligned} $$So the inflection points are:
$$ \begin{matrix} \left( -1, 0 \right) & \left( -0.1748, 2.8311 \right) & \left( 0.7339, 2.0182 \right) & \left( -1.5591, -1.0093 \right)\end{matrix} $$