which statement correctly interprets this limit? lim(x→±∞) (28x - 10x²)/(4x² - 1) the function f(x)=(28x…

Explanation:

Step1: Divide numerator and denominator by $x^{2}$

For $\lim_{x\rightarrow\pm\infty}\frac{28x - 10x^{2}}{4x^{2}-1}$, we get $\lim_{x\rightarrow\pm\infty}\frac{\frac{28x}{x^{2}}-\frac{10x^{2}}{x^{2}}}{\frac{4x^{2}}{x^{2}}-\frac{1}{x^{2}}}=\lim_{x\rightarrow\pm\infty}\frac{\frac{28}{x}- 10}{4-\frac{1}{x^{2}}}$

Step2: Evaluate the limit as $x\rightarrow\pm\infty$

As $x\rightarrow\pm\infty$, $\frac{28}{x}\rightarrow0$ and $\frac{1}{x^{2}}\rightarrow0$. So $\lim_{x\rightarrow\pm\infty}\frac{\frac{28}{x}- 10}{4-\frac{1}{x^{2}}}=\frac{0 - 10}{4-0}=-\frac{5}{2}$ A horizontal - asymptote of a function $y = f(x)$ is given by $y=\lim_{x\rightarrow\pm\infty}f(x)$.

Answer:

A. The function $f(x)=\frac{28x - 10x^{2}}{4x^{2}-1}$ has a horizontal asymptote at $y =-\frac{5}{2}$