Sketch the graph of the polynomial function $$ p(x) = -x^4+x^3+4x^2-x-5 $$

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+x^3+4x^2-x-5 = 0 } $

The solutions of this equation are:

$$ \begin{matrix}x_1 = 1.3356 & x_2 = 2.1406 \end{matrix} $$

(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+x^3+4x^2-x-5 } $, so:

$$ \text{Y inercept} = p(0) = -5 $$

STEP 3:Find the end behavior

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

$$ \lim_{x \to -\infty} \left( -x^4+x^3+4x^2-x-5 \right) = \lim_{x \to -\infty} -x^4 = \color{blue}{ -\infty } $$

The graph starts in the lower-left corner.

$$ \lim_{x \to \infty} \left( -x^4+x^3+4x^2-x-5 \right) = \lim_{x \to \infty} -x^4 = \color{blue}{ -\infty } $$

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) $:

$$ p^{\prime} (x) = -4x^3+3x^2+8x-1 $$

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

$$ p^{\prime} (x) = 0 $$ $$ \begin{matrix}x_1 = 0.1204 & x_2 = -1.16 & x_3 = 1.7895 \end{matrix} $$

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

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

$$ \begin{aligned} \text{for } ~ x & = \color{blue}{ 0.1204 } \Rightarrow p\left(0.1204\right) = \color{orangered}{ -5.0609 }\\[1 em] \text{for } ~ x & = \color{blue}{ -1.16 } \Rightarrow p\left(-1.16\right) = \color{orangered}{ -1.8291 }\\[1 em] \text{for } ~ x & = \color{blue}{ 1.7895 } \Rightarrow p\left(1.7895\right) = \color{orangered}{ 1.4955 }\end{aligned} $$

So the turning points are:

$$ \begin{matrix} \left( 0.1204, -5.0609 \right) & \left( -1.16, -1.8291 \right) & \left( 1.7895, 1.4955 \right)\end{matrix} $$

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+6x+8 $.

The zeros of second derivative are

$$ \begin{matrix}x_1 = 1.1039 & x_2 = -0.6039 \end{matrix} $$

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

$$ \begin{aligned} \text{for } ~ x & = \color{blue}{ 1.1039 } \Rightarrow p\left(1.1039\right) = \color{orangered}{ -1.3692 }\\[1 em] \text{for } ~ x & = \color{blue}{ -0.6039 } \Rightarrow p\left(-0.6039\right) = \color{orangered}{ -3.2905 }\end{aligned} $$

So the inflection points are:

$$ \begin{matrix} \left( 1.1039, -1.3692 \right) & \left( -0.6039, -3.2905 \right)\end{matrix} $$

Graphing Polynomials