Question

determine the vertical asymptotes of the following function. ( f(x) = \frac{2x+6}{x^2 -49} ) the field below accepts a list of numbers or formulas separated by commas (e.g. 2, 4, 6 or ( x + 1, x - 1 )). the order of items does not matter. ( x = square ) check answer

Explanation:

Step1: Factor the denominator

The denominator of the function \( f(x)=\frac{2x + 6}{x^{2}-49} \) is a difference of squares. Recall that \( a^{2}-b^{2}=(a - b)(a + b) \). Here, \( a=x \) and \( b = 7 \), so \( x^{2}-49=(x - 7)(x + 7) \).

Step2: Find values that make denominator zero

To find the vertical asymptotes, we set the denominator equal to zero and solve for \( x \). So we solve the equation \( (x - 7)(x + 7)=0 \). Using the zero - product property (if \( ab = 0 \), then either \( a = 0 \) or \( b = 0 \)), we have \( x-7=0 \) or \( x + 7=0 \). Solving \( x-7=0 \) gives \( x = 7 \), and solving \( x + 7=0 \) gives \( x=-7 \). We also need to check that these values do not make the numerator zero. The numerator is \( 2x+6 \). When \( x = 7 \), \( 2(7)+6=14 + 6=20
eq0 \). When \( x=-7 \), \( 2(-7)+6=-14 + 6=-8
eq0 \). So these values will make the function undefined (since division by zero is undefined) and create vertical asymptotes.

Answer:

\( x=-7,x = 7 \)