Question

given the rational function $\frac{4x^{2}-100}{2x^{2}-7x - 15}$, match each term with its appropriate value or point. horizontal asymptote, y - intercept, x - intercept, vertical asymptote, hole. drag & drop the answer. y = 2, (-5,0), x = 5, (0,6\frac{2}{3}), x=-1.5

Explanation:

Step1: Find horizontal asymptote

For a rational function $\frac{f(x)}{g(x)}$ where $f(x)=4x^{2}-100$ and $g(x)=2x^{2}-7x - 15$, when the degrees of the numerator and denominator are the same (both degree 2), the horizontal asymptote is $y=\frac{a}{b}$, where $a$ is the leading - coefficient of the numerator and $b$ is the leading - coefficient of the denominator. Here, $a = 4$ and $b = 2$, so $y=\frac{4}{2}=2$.

Step2: Find y - intercept

Set $x = 0$ in the function $y=\frac{4x^{2}-100}{2x^{2}-7x - 15}$. Then $y=\frac{4(0)^{2}-100}{2(0)^{2}-7(0)-15}=\frac{- 100}{-15}=\frac{20}{3}=6\frac{2}{3}$, so the y - intercept is $(0,6\frac{2}{3})$.

Step3: Find x - intercept

Set $y = 0$, then $4x^{2}-100 = 0$. Solving for $x$:
\[

$$\begin{align*} 4x^{2}&=100\\ x^{2}&=25\\ x&=\pm5 \end{align*}$$

\]
We need to check if these values make the denominator zero. For the denominator $2x^{2}-7x - 15=(2x + 3)(x - 5)$. When $x = 5$, the denominator is zero, so $x=-5$ is the x - intercept, which is the point $(-5,0)$.

Step4: Find vertical asymptote

Set the denominator $2x^{2}-7x - 15=(2x + 3)(x - 5)=0$. Solving $(2x + 3)(x - 5)=0$ gives $x = 5$ and $x=-\frac{3}{2}=-1.5$.

Step5: Find hole

Factor the numerator $4x^{2}-100 = 4(x + 5)(x - 5)$ and the denominator $2x^{2}-7x - 15=(2x+3)(x - 5)$. Since $(x - 5)$ is a common factor, there is a hole at $x = 5$.

Answer:

Horizontal Asymptote: $y = 2$
Y - Intercept: $(0,6\frac{2}{3})$
X - Intercept: $(-5,0)$
Vertical Asymptote: $x=-1.5,x = 5$
Hole: $x = 5$