Question

8. What is the domain of ="\frac{x}{x^{2} +20x+75}" alt="\frac{x}{x^{2} +20x+75}" align="absmiddle" class="latex-formula"> ? Hint: try factoring the polynomial.
9. What is the domain and range of $\sqrt{13x-7}+1$ ?

Please hurry! I really need help with this!

Answer 1

Answer:

8.  Domain: (-∞, -15) ∪ (-15, -5) ∪ (-5, ∞)

9.  Domain: [7/13, ∞)

    Range: [1, ∞)

Step-by-step explanation:

Question 8

Given rational function:

$f(x)=\dfrac{x}{x^2+20x+75}$

Factor the denominator of the given rational function:

$\implies x^2+20x+75$

$\implies x^2+5x+15x+75$

$\implies x(x+5)+15(x+5)$

$\implies (x+15)(x+5)$

Therefore:

$f(x)=\dfrac{x}{(x+15)(x+5)}$

Asymptote: a line that the curve gets infinitely close to, but never touches.

The function is undefined when the denominator equals zero:

$x+15=0 \implies x=-15$

$x+5=0 \implies x=-5$

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:

$f(x)=\sqrt{13x-7}+1$

Domain: set of all possible input values (x-values)

As the square root of a negative number is undefined:

$\implies 13x-7\geq 0$

$\implies 13x\geq 7$

$\implies x\geq \dfrac{7}{13}$

Therefore, the domain of the given function is:

$\left[\dfrac{7}{13},\infty\right)$

Range: set of all possible output values (y-values)

$\textsf{As }\:\sqrt{13x-7}\geq 0$

$\implies \sqrt{13x-7}+1\geq 1$

Therefore, the range of the given function is:

[1, ∞)