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

Ask by Patel Osborne. in Papua New Guinea
Mar 15,2025

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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 \( y = 0 \).

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To sketch the function \( y=\frac{12}{x-2} \), start by identifying the intercepts. The y-intercept occurs when \( x=0 \): plugging this in gives \( y=\frac{12}{0-2} = -6 \). So, the point \( (0, -6) \) is your y-intercept. The x-intercept occurs when \( y=0 \), but since this fraction never equals zero (the numerator is a constant 12), there is no x-intercept. Next, identify the asymptotes. The vertical asymptote is found where the denominator is zero: \( x - 2 = 0 \) (at \( x=2 \)). The horizontal asymptote can be determined by examining the behavior as \( x \to \pm\infty \): since the value approaches 0, the horizontal asymptote is \( y=0 \). In your sketch, plot the y-intercept at \( (0, -6) \), draw a dashed vertical line at \( x=2 \) for the vertical asymptote, and a dashed horizontal line along \( y=0 \) for the horizontal asymptote. The graph approaches these asymptotes but never touches them, and behaves like two hyperbolic branches, one in the first quadrant and another in the third quadrant. Enjoy sketching those curves!

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