Tap the blue points to see coordinates.
STEP 1:Find the x-intercepts
To find the x-intercepts solve, the equation $ \color{blue}{ x^4-6x^3+3x^2+10x-3 = 0 } $
The solutions of this equation are:
$$ \begin{matrix}x_1 = 0.2887 & x_2 = -1.1371 & x_3 = 1.8157 & x_4 = 5.0326 \end{matrix} $$(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^4-6x^3+3x^2+10x-3 } $, so:
$$ \text{Y inercept} = p(0) = -3 $$STEP 3:Find the end behavior
The end behavior of a polynomial is the same as the end behavior of a leading term.
$$ \lim_{x \to -\infty} \left( x^4-6x^3+3x^2+10x-3 \right) = \lim_{x \to -\infty} x^4 = \color{blue}{ \infty } $$The graph starts in the upper-left corner.
$$ \lim_{x \to \infty} \left( x^4-6x^3+3x^2+10x-3 \right) = \lim_{x \to \infty} x^4 = \color{blue}{ \infty } $$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) $:
$$ p^{\prime} (x) = 4x^3-18x^2+6x+10 $$The x coordinate of the turning points are located at the zeros of the first derivative
$$ p^{\prime} (x) = 0 $$ $$ \begin{matrix}x_1 = -0.5697 & x_2 = 1.1075 & x_3 = 3.9622 \end{matrix} $$(cleck here to see a explanation of how to solve the equation)
To find the y coordinates, substitute the above values into $ p(x) $
$$ \begin{aligned} \text{for } ~ x & = \color{blue}{ -0.5697 } \Rightarrow p\left(-0.5697\right) = \color{orangered}{ -6.5086 }\\[1 em] \text{for } ~ x & = \color{blue}{ 1.1075 } \Rightarrow p\left(1.1075\right) = \color{orangered}{ 5.1086 }\\[1 em] \text{for } ~ x & = \color{blue}{ 3.9622 } \Rightarrow p\left(3.9622\right) = \color{orangered}{ -43.0376 }\end{aligned} $$So the turning points are:
$$ \begin{matrix} \left( -0.5697, -6.5086 \right) & \left( 1.1075, 5.1086 \right) & \left( 3.9622, -43.0376 \right)\end{matrix} $$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^2-36x+6 $.
The zeros of second derivative are
$$ \begin{matrix}x_1 = 2.8229 & x_2 = 0.1771 \end{matrix} $$Substitute the x values into $ p(x) $ to get y coordinates
$$ \begin{aligned} \text{for } ~ x & = \color{blue}{ 2.8229 } \Rightarrow p\left(2.8229\right) = \color{orangered}{ -22.333 }\\[1 em] \text{for } ~ x & = \color{blue}{ 0.1771 } \Rightarrow p\left(0.1771\right) = \color{orangered}{ -1.167 }\end{aligned} $$So the inflection points are:
$$ \begin{matrix} \left( 2.8229, -22.333 \right) & \left( 0.1771, -1.167 \right)\end{matrix} $$