Answer:
8. Domain: (-∞, -15) ∪ (-15, -5) ∪ (-5, ∞)
9. Domain: [7/13, ∞)
Range: [1, ∞)
Step-by-step explanation:
Question 8
Given rational function:
[tex]f(x)=\dfrac{x}{x^2+20x+75}[/tex]
Factor the denominator of the given rational function:
[tex]\implies x^2+20x+75[/tex]
[tex]\implies x^2+5x+15x+75[/tex]
[tex]\implies x(x+5)+15(x+5)[/tex]
[tex]\implies (x+15)(x+5)[/tex]
Therefore:
[tex]f(x)=\dfrac{x}{(x+15)(x+5)}[/tex]
Asymptote: a line that the curve gets infinitely close to, but never touches.
The function is undefined when the denominator equals zero:
[tex]x+15=0 \implies x=-15[/tex]
[tex]x+5=0 \implies x=-5[/tex]
Therefore, there are vertical asymptotes at x = -15 and x = -5.
Domain: set of all possible input values (x-values)
Therefore, the domain of the given rational function is:
(-∞, -15) ∪ (-15, -5) ∪ (-5, ∞)
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Question 9
Given function:
[tex]f(x)=\sqrt{13x-7}+1[/tex]
Domain: set of all possible input values (x-values)
As the square root of a negative number is undefined:
[tex]\implies 13x-7\geq 0[/tex]
[tex]\implies 13x\geq 7[/tex]
[tex]\implies x\geq \dfrac{7}{13}[/tex]
Therefore, the domain of the given function is:
[tex]\left[\dfrac{7}{13},\infty\right)[/tex]
Range: set of all possible output values (y-values)
[tex]\textsf{As }\:\sqrt{13x-7}\geq 0[/tex]
[tex]\implies \sqrt{13x-7}+1\geq 1[/tex]
Therefore, the range of the given function is:
[1, ∞)
