Question Five Sketch the function $$ y=\frac{12}{x-2} $$ clearly showing the intercepts and asymptotes $$ \quad[5 \mathrm{mks}] $$

Question

UpStudy AI · Accepted Answer

1. **Identify the domain:** The function is $$ y = \frac{12}{x-2}. $$ The denominator cannot be zero, hence $$ x - 2 \neq 0 \quad \Rightarrow \quad x \neq 2. $$ Therefore, the domain is all real numbers except $$ x = 2 $$. 2. **Find the vertical asymptote:** The vertical asymptote occurs when the denominator is zero. Since $$ x - 2 = 0 $$ when $$ x = 2 $$, there is a vertical asymptote at $$ x = 2. $$ 3. **Find the horizontal asymptote:** For rational functions, compare the degrees of the numerator and denominator. The numerator $$ 12 $$ is of degree 0, and the denominator $$ x-2 $$ is of degree 1. Since the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is $$ y = 0. $$ 4. **Determine the intercepts:** - **Y-intercept:** Set $$ x = 0 $$ in the equation: $$ y = \frac{12}{0-2} = \frac{12}{-2} = -6. $$ Thus, the y-intercept is $$ (0, -6) $$. - **X-intercept:** Set $$ y = 0 $$: $$ \frac{12}{x-2} = 0. $$ A fraction equals zero only when its numerator is zero. Since the numerator is $$ 12 $$ (which is never $$ 0 $$), there is no x-intercept. 5. **Sketching and behavior analysis:** - The vertical asymptote is $$ x = 2 $$ and the horizontal asymptote is $$ y = 0 $$. - For $$ x > 2 $$: The denominator $$ (x-2) $$ is positive, so $$ y $$ is positive. As $$ x \to 2^+ $$, $$ y \to +\infty $$; and as $$ x \to +\infty $$, $$ y \to 0^+ $$. - For $$ x < 2 $$: The denominator $$ (x-2) $$ is negative, so $$ y $$ is negative. As $$ x \to 2^- $$, $$ y \to -\infty $$; and as $$ x \to -\infty $$, $$ y \to 0^- $$. The graph consists of two branches of a hyperbola: - The branch for $$ x > 2 $$ lies in the first quadrant (approaching the horizontal asymptote $$ y=0 $$ from above) and rises steeply near $$ x=2 $$. - The branch for $$ x < 2 $$ lies in the third quadrant (approaching $$ y=0 $$ from below) and falls sharply near the vertical asymptote $$ x=2 $$. These steps provide all necessary details to sketch the function $$ y=\frac{12}{x-2} $$ with intercepts and asymptotes. To sketch the function $$ y = \frac{12}{x-2} $$: 1. **Domain:** All real numbers except $$ x = 2 $$. 2. **Vertical Asymptote:** $$ x = 2 $$. 3. **Horizontal Asymptote:** $$ y = 0 $$. 4. **Intercepts:** - **Y-intercept:** $$ (0, -6) $$. - **No X-intercept.** 5. **Behavior:** - For $$ x > 2 $$: $$ y $$ is positive and approaches $$ +\infty $$ as $$ x $$ approaches 2 from the right, and approaches 0 as $$ x $$ increases. - For $$ x < 2 $$: $$ y $$ is negative and approaches $$ -\infty $$ as $$ x $$ approaches 2 from the left, and approaches 0 as $$ x $$ decreases. The graph consists of two branches of a hyperbola with the vertical asymptote at $$ x = 2 $$ and horizontal asymptote at $$ y = 0 $$.

Question Five Sketch the function \( y=\frac{12}{x-2} \) clearly showing the intercepts and asymptotes \( \quad[5 \mathrm{mks}] \)

Upstudy AI Solution

Answer

Solution

Beyond the Answer

Related Questions

Latest Pre Calculus Questions