Sketch the graph of the polynomial function $$ p(x) = x^2+2x^2-x-2 $$

Tap the blue points to see coordinates.

STEP 1:Find the x-intercepts

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

The solutions of this equation are:

$$ \begin{matrix}x_1 = 1 & x_2 = -\dfrac{ 2 }{ 3 } \end{matrix} $$

(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^2+2x^2-x-2 } $, so:

$$ \text{Y inercept} = p(0) = -2 $$

STEP 3:Find the end behavior

The end behavior of a polynomial is the same as the end behavior of a leading term.

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

The graph starts in the upper-left corner.

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

The graph ends in the upper-right corner.

STEP 4:Find the turning points

To determine the turning point, we need to find the first derivative of $ p(x) $:

$$ p^{\prime} (x) = 6x-1 $$

The x coordinate of the turning point is located at the zeros of the first derivative

$$ p^{\prime} (x) = 0 $$ $$ \begin{matrix}x = \dfrac{ 1 }{ 6 } \end{matrix} $$

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

To find the y coordinate, substitute the above value into $ p(x) $

$$ \begin{aligned} \text{for } ~ x & = \color{blue}{ \frac{ 1 }{ 6 } } \Rightarrow p\left(\frac{ 1 }{ 6 }\right) = \color{orangered}{ -\frac{ 25 }{ 12 } }\end{aligned} $$

So the turning point is:

$$ \begin{matrix} \left( \dfrac{ 1 }{ 6 }, -\dfrac{ 25 }{ 12 } \right)\end{matrix} $$

Graphing Polynomials