Find the domain, range and sketch the graph of the function f(x)= [2−x]/{4x−16] through the asymptotes and the x,y-intercept points.
The Correct Answer and Explanation is:
To analyze the function f(x)=2−x4x−16f(x) = \frac{2 – x}{4x – 16}f(x)=4x−162−x, we will determine its domain, range, asymptotes, intercepts, and graph it.
Domain:
The denominator 4x−16=04x – 16 = 04x−16=0 when x=4x = 4x=4. At x=4x = 4x=4, the function is undefined.
Domain: x∈(−∞,4)∪(4,∞)x \in (-\infty, 4) \cup (4, \infty)x∈(−∞,4)∪(4,∞).
Intercepts:
- x-intercept: The numerator equals 0 when 2−x=02 – x = 02−x=0, i.e., x=2x = 2x=2.
Thus, the x-intercept is (2,0)(2, 0)(2,0). - y-intercept: Substituting x=0x = 0x=0,f(0)=2−04(0)−16=2−16=−18.f(0) = \frac{2 – 0}{4(0) – 16} = \frac{2}{-16} = -\frac{1}{8}.f(0)=4(0)−162−0=−162=−81.Thus, the y-intercept is (0,−18)(0, -\frac{1}{8})(0,−81).
Asymptotes:
- Vertical asymptote: The denominator 4x−16=04x – 16 = 04x−16=0, so the vertical asymptote is x=4x = 4x=4.
- Horizontal asymptote: As x→±∞x \to \pm\inftyx→±∞, the highest degree terms dominate. f(x)∼−x4x=−14.f(x) \sim \frac{-x}{4x} = -\frac{1}{4}.f(x)∼4x−x=−41. Thus, the horizontal asymptote is y=−14y = -\frac{1}{4}y=−41.
Range:
The function approaches −14-\frac{1}{4}−41 but never equals it, and spans all other values because there are no restrictions on f(x)f(x)f(x).
Range: y∈(−∞,−14)∪(−14,∞)y \in (-\infty, -\frac{1}{4}) \cup (-\frac{1}{4}, \infty)y∈(−∞,−41)∪(−41,∞).
Behavior and Sketch:
- Near x=4x = 4x=4, f(x)f(x)f(x) diverges to ±∞\pm\infty±∞ depending on the sign of x−4x – 4x−4.
- For large xxx, f(x)f(x)f(x) stabilizes near y=−14y = -\frac{1}{4}y=−41.
The graph shows a hyperbolic shape, with the x-intercept at (2,0)(2, 0)(2,0), y-intercept at (0,−18)(0, -\frac{1}{8})(0,−81), and asymptotes at x=4x = 4x=4 and y=−14y = -\frac{1}{4}y=−41.
Would you like me to generate the graph?