7. let h be the function defined by h(x)=\frac{7x^{2}+14x}{x^{2}-4}. answer the following questions: a…

7. let h be the function defined by h(x)=\frac{7x^{2}+14x}{x^{2}-4}. answer the following questions: a. lim_{x\rightarrow - 2}h(x)= b. what is the vertical asymptote? c. at what x - value is there a removable discontinuity?

7. let h be the function defined by h(x)=\frac{7x^{2}+14x}{x^{2}-4}. answer the following questions: a. lim_{x\rightarrow - 2}h(x)= b. what is the vertical asymptote? c. at what x - value is there a removable discontinuity?

Answer

Explanation:

Step1: Simplify the function

First, factor the numerator and denominator of (h(x)=\frac{7x^{2}+14x}{x^{2}-4}). The numerator (7x^{2}+14x = 7x(x + 2)) and the denominator (x^{2}-4=(x + 2)(x - 2)). So (h(x)=\frac{7x(x + 2)}{(x + 2)(x - 2)}=\frac{7x}{x - 2},x\neq - 2).

Step2: Find the vertical - asymptote

A vertical asymptote occurs where the denominator of a rational function is zero and the numerator is non - zero. Set the denominator of the simplified function (y=\frac{7x}{x - 2}) equal to zero. Solving (x-2 = 0), we get (x = 2).

Step3: Find the removable discontinuity

A removable discontinuity occurs when a factor can be canceled out in the numerator and denominator. Since we canceled out the factor ((x + 2)) in the simplification process, the function has a removable discontinuity at (x=-2).

Answer:

The vertical asymptote is (x = 2) and the (x) - value of the removable discontinuity is (x=-2).