STEP 1:Find the x-intercepts

To find the x-intercepts solve, the equation $ \color{blue}{ -3x^4+x^3+x^2 = 0 } $

The solutions of this equation are:

$$ \begin{matrix}x_1 = 0 & x_2 = 0.7676 & x_3 = -0.4343 \end{matrix} $$

(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) = -3x^4+x^3+x^2 } $, 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.

$$ \lim_{x \to -\infty} \left( -3x^4+x^3+x^2 \right) = \lim_{x \to -\infty} -3x^4 = \color{blue}{ -\infty } $$

The graph starts in the lower-left corner.

$$ \lim_{x \to \infty} \left( -3x^4+x^3+x^2 \right) = \lim_{x \to \infty} -3x^4 = \color{blue}{ -\infty } $$

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) $:

$$ p^{\prime} (x) = -12x^3+3x^2+2x $$

The x coordinate of the turning points are located at the zeros of the first derivative

$$ p^{\prime} (x) = 0 $$ $$ \begin{matrix}x_1 = 0 & x_2 = 0.552 & x_3 = -0.302 \end{matrix} $$

(cleck here to see a explanation of how to solve the equation)

To find the y coordinates, substitute the above values into $ p(x) $

$$ \begin{aligned} \text{for } ~ x & = \color{blue}{ 0 } \Rightarrow p\left(0\right) = \color{orangered}{ 0 }\\[1 em] \text{for } ~ x & = \color{blue}{ 0.552 } \Rightarrow p\left(0.552\right) = \color{orangered}{ 0.1944 }\\[1 em] \text{for } ~ x & = \color{blue}{ -0.302 } \Rightarrow p\left(-0.302\right) = \color{orangered}{ 0.0387 }\end{aligned} $$

So the turning points are:

$$ \begin{matrix} \left( 0, 0 \right) & \left( 0.552, 0.1944 \right) & \left( -0.302, 0.0387 \right)\end{matrix} $$

STEP 5:Find the inflection points

The inflection points are located at zeroes of second derivative. The second derivative is $ p^{\prime \prime} (x) = -36x^2+6x+2 $.

The zeros of second derivative are

$$ \begin{matrix}x_1 = \dfrac{ 1 }{ 3 } & x_2 = -\dfrac{ 1 }{ 6 } \end{matrix} $$

Substitute the x values into $ p(x) $ to get y coordinates

$$ \begin{aligned} \text{for } ~ x & = \color{blue}{ \frac{ 1 }{ 3 } } \Rightarrow p\left(\frac{ 1 }{ 3 }\right) = \color{orangered}{ \frac{ 1 }{ 9 } }\\[1 em] \text{for } ~ x & = \color{blue}{ -\frac{ 1 }{ 6 } } \Rightarrow p\left(-\frac{ 1 }{ 6 }\right) = \color{orangered}{ \frac{ 1 }{ 48 } }\end{aligned} $$

So the inflection points are:

$$ \begin{matrix} \left( \dfrac{ 1 }{ 3 }, \dfrac{ 1 }{ 9 } \right) & \left( -\dfrac{ 1 }{ 6 }, \dfrac{ 1 }{ 48 } \right)\end{matrix} $$

Graphing Polynomials