f(x)=\frac{-2x^{2}+14}{x^{2}-49}\nwhich statement describes the behavior of the graph of the function shown…

f(x)=\frac{-2x^{2}+14}{x^{2}-49}\nwhich statement describes the behavior of the graph of the function shown at the vertical asymptotes?\nas x approaches -7 from the left, y approaches (infty).\nas x approaches -7 from the right, y approaches (-infty).\nas x approaches 7 from the left, y approaches (-infty).\nas x approaches 7 from the right, y approaches (-infty).

f(x)=\frac{-2x^{2}+14}{x^{2}-49}\nwhich statement describes the behavior of the graph of the function shown at the vertical asymptotes?\nas x approaches -7 from the left, y approaches (infty).\nas x approaches -7 from the right, y approaches (-infty).\nas x approaches 7 from the left, y approaches (-infty).\nas x approaches 7 from the right, y approaches (-infty).

Answer

Answer:

C. As x approaches 7 from the left, y approaches $-\infty$.

Explanation:

Step1: Find vertical asymptotes

Set denominator $x^{2}-49 = 0$. Then $(x + 7)(x - 7)=0$, so $x=-7$ and $x = 7$ are vertical asymptotes.

Step2: Analyze $x\to - 7$

For $x\to - 7$, consider $\lim_{x\to - 7^{-}}\frac{-2x^{2}+14}{x^{2}-49}$ and $\lim_{x\to - 7^{+}}\frac{-2x^{2}+14}{x^{2}-49}$. When $x\to - 7^{-}$, numerator is negative and denominator is negative, so $y\to+\infty$. When $x\to - 7^{+}$, numerator is negative and denominator is positive, so $y\to-\infty$.

Step3: Analyze $x\to7$

For $x\to7$, consider $\lim_{x\to7^{-}}\frac{-2x^{2}+14}{x^{2}-49}$ and $\lim_{x\to7^{+}}\frac{-2x^{2}+14}{x^{2}-49}$. When $x\to7^{-}$, numerator is negative and denominator is negative, so $y\to-\infty$. When $x\to7^{+}$, numerator is negative and denominator is positive, so $y\to-\infty$.