Q.15 - Precalculus Enhanced with Graphing Utilities

Question

(a) Find a polynomial function of degree 3 whose y-intercept

is 5 and whose x-intercepts are -2, 3, and 5. Graph the

function.

(b) Find a rational function whose y-intercept is 5 and

whose x-intercepts are -2, 3, and 5 that has the line

x = 2 as a vertical asymptote. Graph the function.

Step-by-Step Solution

Answer

part (a) (1) The polynomial function is :$6y=x^3-6x^2-x+30$

(2) Graph of function is :

part(b)(1) The function is :$y=\frac{-{\displaystyle \frac{1}{3}}{x}^3+2x^2+{\displaystyle \frac{1}{3}}x-10}{x-2}$

 

(2) The graph of function is :

part(a)step1. Given information

Intercepts of the curve on 

the x-axis (-2,3,5)

the y axis (5)

Now,

 let's assume a 3-degree polynomial

$y=a{x}^3+b{x}^2+cx+d$

Step2. Getting the value the unknown

as y intersects is 5

so

the x intersects are$-2,3,5$ 

so

$$\begin{gathered} 0={\left(-2\right)}^{3}a+b{\left(-2\right)}^2+\left(-2\right)c+5 \\ 0=-8a+4b-2c+5 \end{gathered}$$

for x= 3

$$\begin{gathered} 0={\left(3\right)}^{3}a+{\left(3\right)}^{2}b+3c+5 \\ 0= 27a+9b+3c+5 \end{gathered}$$

for x= 5

$$\begin{gathered} 0={\left(5\right)}^{3}a+{\left(5\right)}^{2}b+5c+5 \\ 0=125a+25b+5c+5 \end{gathered}$$

solving these 

$a=\frac{1}{6}, b=-1, c=\frac{-1}{6}$

and equation is 

$6y= x^3-6x^2-x+30$

part(b) step1. assume the function

We are given that the line x=2

Is the vertical asymptote of the rational function 

Therefore the denominator of the ration function is 

X-2

Since the function has three intersects points on the x-axis 

So 

Let's assume the rational function as

$y= \frac{a{x}^3+b{x}^2+cx+d}{x-2}$

step 4. calculating the value of unknown

the y intersects point is 5

so 

$$\begin{gathered} 5=\frac{d}{-2} \\ d=-10 \end{gathered}$$

the x intersects points are 2 and 5 so

$$\begin{gathered} 0= {\left(-2\right)}^{3}a+{\left(-2\right)}^{2}b+\left(-2\right)c-10 \\ 8a-4b+2c+10=0 \end{gathered}$$

and$$\begin{gathered} 0={\left(5\right)}^{3}a+{\left(5\right)}^{2}b+\left(5\right)c-10 \\ 10= 125a+25b+5c \end{gathered}$$

step5.solving equation

$a=\frac{-1}{3}, b=2,c=\frac{1}{3}$