Sketch the graph of the polynomial function $$ p(x) = 9x^4-30x^3+13x^2+20+4 $$

Tap the blue points to see coordinates.

STEP 1:Find the x-intercepts

To find the x-intercepts solve, the equation $ \color{blue}{ 9x^4-30x^3+13x^2+20+4 = 0 } $

The solutions of this equation are:

$$ \begin{matrix}x_1 = 1.4418 & x_2 = 2.642 \end{matrix} $$

(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) = 9x^4-30x^3+13x^2+20+4 } $, so:

$$ \text{Y inercept} = p(0) = 24 $$

STEP 3:Find the end behavior

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

$$ \lim_{x \to -\infty} \left( 9x^4-30x^3+13x^2+20+4 \right) = \lim_{x \to -\infty} 9x^4 = \color{blue}{ \infty } $$

The graph starts in the upper-left corner.

$$ \lim_{x \to \infty} \left( 9x^4-30x^3+13x^2+20+4 \right) = \lim_{x \to \infty} 9x^4 = \color{blue}{ \infty } $$

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) $:

$$ p^{\prime} (x) = 36x^3-90x^2+26x $$

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

$$ p^{\prime} (x) = 0 $$ $$ \begin{matrix}x_1 = 0 & x_2 = \dfrac{ 13 }{ 6 } & x_3 = \dfrac{ 1 }{ 3 } \end{matrix} $$

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

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

$$ \begin{aligned} \text{for } ~ x & = \color{blue}{ 0 } \Rightarrow p\left(0\right) = \color{orangered}{ 24 }\\[1 em] \text{for } ~ x & = \color{blue}{ \frac{ 13 }{ 6 } } \Rightarrow p\left(\frac{ 13 }{ 6 }\right) = \color{orangered}{ -\frac{ 1045 }{ 48 } }\\[1 em] \text{for } ~ x & = \color{blue}{ \frac{ 1 }{ 3 } } \Rightarrow p\left(\frac{ 1 }{ 3 }\right) = \color{orangered}{ \frac{ 220 }{ 9 } }\end{aligned} $$

So the turning points are:

$$ \begin{matrix} \left( 0, 24 \right) & \left( \dfrac{ 13 }{ 6 }, -\dfrac{ 1045 }{ 48 } \right) & \left( \dfrac{ 1 }{ 3 }, \dfrac{ 220 }{ 9 } \right)\end{matrix} $$

STEP 5:Find the inflection points

The inflection points are located at zeroes of second derivative. The second derivative is $ p^{\prime \prime} (x) = 108x^2-180x+26 $.

The zeros of second derivative are

$$ \begin{matrix}x_1 = 1.5069 & x_2 = 0.1598 \end{matrix} $$

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

$$ \begin{aligned} \text{for } ~ x & = \color{blue}{ 1.5069 } \Rightarrow p\left(1.5069\right) = \color{orangered}{ -2.7277 }\\[1 em] \text{for } ~ x & = \color{blue}{ 0.1598 } \Rightarrow p\left(0.1598\right) = \color{orangered}{ 24.2153 }\end{aligned} $$

So the inflection points are:

$$ \begin{matrix} \left( 1.5069, -2.7277 \right) & \left( 0.1598, 24.2153 \right)\end{matrix} $$

Graphing Polynomials