Sketch the graph of the polynomial function $$ p(x) = x^3+4x^2-x-7 $$

Tap the blue points to see coordinates.

STEP 1:Find the x-intercepts

To find the x-intercepts solve, the equation $ \color{blue}{ x^3+4x^2-x-7 = 0 } $

The solutions of this equation are:

$$ \begin{matrix}x_1 = 1.2534 & x_2 = -1.48 & x_3 = -3.7734 \end{matrix} $$

(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^3+4x^2-x-7 } $, so:

$$ \text{Y inercept} = p(0) = -7 $$

STEP 3:Find the end behavior

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

$$ \lim_{x \to -\infty} \left( x^3+4x^2-x-7 \right) = \lim_{x \to -\infty} x^3 = \color{blue}{ -\infty } $$

The graph starts in the lower-left corner.

$$ \lim_{x \to \infty} \left( x^3+4x^2-x-7 \right) = \lim_{x \to \infty} x^3 = \color{blue}{ \infty } $$

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) $:

$$ p^{\prime} (x) = 3x^2+8x-1 $$

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

$$ p^{\prime} (x) = 0 $$ $$ \begin{matrix}x_1 = 0.1196 & x_2 = -2.7863 \end{matrix} $$

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

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

$$ \begin{aligned} \text{for } ~ x & = \color{blue}{ 0.1196 } \Rightarrow p\left(0.1196\right) = \color{orangered}{ -7.0607 }\\[1 em] \text{for } ~ x & = \color{blue}{ -2.7863 } \Rightarrow p\left(-2.7863\right) = \color{orangered}{ 5.2088 }\end{aligned} $$

So the turning points are:

$$ \begin{matrix} \left( 0.1196, -7.0607 \right) & \left( -2.7863, 5.2088 \right)\end{matrix} $$

STEP 5:Find the inflection points

The inflection points are located at zeroes of second derivative. The second derivative is $ p^{\prime \prime} (x) = 6x+8 $.

The zero of second derivative is

$$ \begin{matrix}x = -\dfrac{ 4 }{ 3 } \end{matrix} $$

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

$$ \begin{aligned} \text{for } ~ x & = \color{blue}{ -\frac{ 4 }{ 3 } } \Rightarrow p\left(-\frac{ 4 }{ 3 }\right) = \color{orangered}{ -\frac{ 25 }{ 27 } }\end{aligned} $$

So the inflection point is:

$$ \begin{matrix} \left( -\dfrac{ 4 }{ 3 }, -\dfrac{ 25 }{ 27 } \right)\end{matrix} $$

Graphing Polynomials