The general equation of line is :
[tex]y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)[/tex]The equation of the function g(x) is:
Consider any two coordinates : (-3,0) & (0,3) So, the equation is:
[tex]\begin{gathered} y-0=\frac{3-0}{0-(-3)}(x-(-3)) \\ y=\frac{3}{3}(x+3) \\ y=x+3 \\ g(x)=x+3 \end{gathered}[/tex]The equation of the function f(x) is:
Consider any two coordinates: (-1,0) & (0,1)
[tex]\begin{gathered} y-0=\frac{1-0}{0-(-1)}(x-(-1)) \\ y=1(x+1)_{} \\ y=x+1 \\ f(x)=x+1 \end{gathered}[/tex]Since, h(x) = f(x)/g(x)
So, the funstion h(x) is express as:
[tex]\begin{gathered} h(x)=\frac{f(x)}{g(x)} \\ h(x)=\frac{x+1}{x+3} \end{gathered}[/tex]a) Vertical intercept of h
Substitute x =0 in h(x)
[tex]\begin{gathered} h(x)=\frac{x+1}{x+3} \\ h(0)=\frac{0+1}{0+3} \\ h(0)=\frac{1}{3} \end{gathered}[/tex]b) Determine the roots of h(x)
[tex]\begin{gathered} g(x)=\frac{x+1}{x+3} \\ \text{ the given expression is the iraationla polynomial } \\ \frac{x+1}{x+3}=\frac{x+1}{x+3} \\ \text{ Roots of h(x) =}\frac{x+1}{x+3} \end{gathered}[/tex]c) Vertical asymptote : Vertical asymptotes are vertical lines which correspond to the zeroes of the denominator of a rational function
So,
[tex]\begin{gathered} x+3=0 \\ x=-3 \end{gathered}[/tex]The vertical assymptote : x =-3
D) Horizontal assymptote : x - 3
bAnswer: x -3
Answer:
a)h(0)=1/3
b) h(x)=(x+1)/(x+3)
c) x = -3
Hroizontal Assymotote : y =1/3
