Find the horizontal intercepts, the vertical intercept, the vertical asymptotes, and the horizontal or slant asymptote of the function. Use that information to sketch a graph.
f(x)=x^2−x−20/x^2−16
Enter the intercepts as points, (a,b).
The x-intercept is_____________
The y-intercept is__________
The field below accepts a list of numbers or formulas separated by semicolons (e.g. 2;4;6 or x+1;x−1). The order of the list does not matter.
Vertical asymptotes:
x=
Horizontal or slant asymptote:
y=
The Correct Answer and Explanation is :
To analyze the function f(x)=x2−x−20×2−16f(x) = \frac{x^2 – x – 20}{x^2 – 16}, we’ll determine its intercepts and asymptotes, and then sketch its graph.
1. Vertical Asymptotes:
Vertical asymptotes occur where the denominator is zero, provided the numerator isn’t zero at those points.
The denominator is x2−16=(x−4)(x+4)x^2 – 16 = (x – 4)(x + 4). Setting it to zero gives:
x2−16=0 ⟹ x=±4x^2 – 16 = 0 \implies x = \pm 4
At x=4x = 4:
f(4)=42−4−2042−16=16−4−2016−16=−80 ⟹ undefinedf(4) = \frac{4^2 – 4 – 20}{4^2 – 16} = \frac{16 – 4 – 20}{16 – 16} = \frac{-8}{0} \implies \text{undefined}
At x=−4x = -4:
f(−4)=(−4)2+4−20(−4)2−16=16+4−2016−16=00 ⟹ indeterminatef(-4) = \frac{(-4)^2 + 4 – 20}{(-4)^2 – 16} = \frac{16 + 4 – 20}{16 – 16} = \frac{0}{0} \implies \text{indeterminate}
Since both points lead to division by zero, x=4x = 4 and x=−4x = -4 are vertical asymptotes.
2. Horizontal Asymptote:
For rational functions, if the degrees of the numerator and denominator are equal, the horizontal asymptote is the ratio of their leading coefficients.
Here, both numerator and denominator are degree 2, with leading coefficients 1. Thus, the horizontal asymptote is:
y=11=1y = \frac{1}{1} = 1
3. Intercepts:
- Y-Intercept: Set x=0x = 0: f(0)=02−0−2002−16=−20−16=54f(0) = \frac{0^2 – 0 – 20}{0^2 – 16} = \frac{-20}{-16} = \frac{5}{4} So, the y-intercept is (0,54)\left(0, \frac{5}{4}\right).
- X-Intercepts: Set the numerator equal to zero: x2−x−20=0x^2 – x – 20 = 0 Factoring: (x−5)(x+4)=0(x – 5)(x + 4) = 0 So, x=5x = 5 and x=−4x = -4. However, x=−4x = -4 is a vertical asymptote, not an intercept. Therefore, the x-intercept is (5,0)(5, 0).
Summary:
- X-Intercept: (5,0)(5, 0)
- Y-Intercept: (0,54)\left(0, \frac{5}{4}\right)
- Vertical Asymptotes: x=4x = 4 and x=−4x = -4
- Horizontal Asymptote: y=1y = 1
Graph Sketch:
The function has vertical asymptotes at x=4x = 4 and x=−4x = -4, causing the graph to approach infinity near these lines. The horizontal asymptote at y=1y = 1 indicates that as xx approaches ±∞, the function approaches 1. The graph crosses the x-axis at (5,0)(5, 0) and the y-axis at (0,54)\left(0, \frac{5}{4}\right).
Here’s a sketch of the graph:
“`plaintext y | 4| * * * 3| \ / \ / 2| \ / \ / 1|—————-/——–/————- 0| */ */ x -1| / \ /
-2| / \ /
-3| / \ /
-4| * * * |
*Note: The asterisks (*) represent the vertical asymptotes at \( x = -4 \) and \( x = 4 \). The graph approaches these lines but never touches them.*