Question

let h be the function defined by h(x) = (7x²+14x)/(x² - 4). answer the following questions. a) at what x - value is there a removable discontinuity? x = - 2 b) what is the vertical asymptote? x = 2 c) lim(x→ - 2) h(x) = - 7

Explanation:

Step1: Factor the function

Given \(h(x)=\frac{7x^{2}+14x}{x^{2}-4}\), factor the numerator and denominator. Numerator: \(7x^{2}+14x = 7x(x + 2)\), denominator: \(x^{2}-4=(x + 2)(x - 2)\). So \(h(x)=\frac{7x(x + 2)}{(x + 2)(x - 2)}\), \(x
eq\pm2\).

Step2: Find removable discontinuity

A removable discontinuity occurs when a factor in the numerator and denominator cancels out. Here, the factor \((x + 2)\) cancels. The \(x\) - value for the removable discontinuity is \(x=-2\) since when \(x=-2\), the original function is undefined but the simplified - function (after canceling) is well - defined.

Step3: Find vertical asymptote

A vertical asymptote occurs at the \(x\) - values that make the denominator of the simplified non - canceled function equal to zero. Set \(x-2 = 0\), then \(x = 2\).

Answer:

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