Math Problem Statement

Sketch the graph of an example of a function f that satisfies all of the given conditions. f(0) = 0, lim x→2− (f(x)) = ∞, lim x→2+ (f(x)) = −∞, lim x→−∞ (f(x)) = −4, lim x→∞ (f(x)) = −4 choose the correct graph a. The x y-coordinate plane is given. A curve, a vertical dashed line, and a horizontal dashed line are graphed. A vertical dashed line crosses the x-axis at x = 2. A horizontal dashed line crosses the y-axis at y = −4. The curve with 2 parts enters the window almost horizontally above y = −4, goes up and right becoming more steep, passes through the origin, exits almost vertically just to the left of x = 2, reenters almost vertically just to the right of x = 2, goes up and right becoming less steep, and exits the window almost horizontally below y = −4. b. The x y-coordinate plane is given. A curve, 2 vertical dashed lines, and a horizontal dashed line are graphed. A vertical dashed line crosses the x-axis at x = −4. A vertical dashed line crosses the x-axis at x = 4. A horizontal dashed line crosses the y-axis at y = 2. The curve in 3 parts enters the window almost horizontally above y = 2, goes up and right becoming more steep, exits almost vertically just to the left of x = −4, reenters almost vertically just to the right of x = −4, goes up and right becoming less steep, changes directions at the origin, goes down and right becoming more steep, exits almost vertically just to the left of x = 4, reenters almost vertically just to the right of x = 4, goes down and right becoming less steep, and exits the window almost horizontally above y = 2. c. The x y-coordinate plane is given. A curve, 2 vertical dashed lines, and a horizontal dashed line are graphed. A vertical dashed line crosses the x-axis at x = −4. A vertical dashed line crosses the x-axis at x = 4. A horizontal dashed line crosses the y-axis at y = 2. The curve with 3 parts enters the window almost horizontally above y = 2, goes up and right becoming more steep, exits almost vertically just to the left of x = −4, reenters almost vertically just to the right of x = −4, goes up and right becoming less steep, crosses the x-axis at approximately x = −3.9, changes direction at the approximate point (0, 1.9), goes down and right becoming more steep, crosses the x-axis at approximately x = 3.9, exits almost vertically just to the left of x = 4, reenters almost vertically just to the right of x = 4, goes down and right becoming less steep, and exits the window almost horizontally above y = 2. d. The x y-coordinate plane is given. A curve, a vertical dashed line, and a horizontal dashed line are graphed. A vertical dashed line crosses the x-axis at x = 2. A horizontal dashed line crosses the y-axis at y = −4. The curve with 2 parts enters the window almost horizontally above y = −4, goes up and right becoming more steep, passes through the origin, exits almost vertically just to the left of x = 2, reenters almost vertically just to the right of x = 2, goes down and right becoming less steep, crosses the x-axis at approximately x = 2.5, and exits the window almost horizontally above y = −4.

Solution

To determine the correct graph, let's analyze the given conditions one by one:

  1. f(0)=0f(0) = 0: This indicates that the function passes through the origin, i.e., the point (0, 0).

  2. lim⁡x→2−f(x)=∞\lim_{x \to 2^-} f(x) = \infty: As xx approaches 2 from the left, the function tends to positive infinity. This suggests that the graph has a vertical asymptote at x=2x = 2 from the left, where the curve goes upward infinitely.

  3. lim⁡x→2+f(x)=−∞\lim_{x \to 2^+} f(x) = -\infty: As xx approaches 2 from the right, the function tends to negative infinity. So, the graph approaches negative infinity to the right of x=2x = 2.

  4. lim⁡x→−∞f(x)=−4\lim_{x \to -\infty} f(x) = -4: As xx tends towards negative infinity, the function approaches the horizontal asymptote y=−4y = -4.

  5. lim⁡x→∞f(x)=−4\lim_{x \to \infty} f(x) = -4: Similarly, as xx tends towards positive infinity, the function approaches the horizontal asymptote y=−4y = -4.

Now, let's match these conditions with the graphs:

  • Graph A:
    • It has a vertical asymptote at x=2x = 2, and the curve goes to ∞\infty from the left and −∞-\infty from the right.
    • The graph passes through the origin.
    • The curve approaches y=−4y = -4 as x→−∞x \to -\infty and x→∞x \to \infty.

This graph satisfies all the given conditions.

  • Graph B:

    • It has vertical asymptotes at x=−4x = -4 and x=4x = 4, which does not match the condition that there should be a vertical asymptote at x=2x = 2.

  • Graph C:

    • Similar to Graph B, it has vertical asymptotes at x=−4x = -4 and x=4x = 4, which does not match the conditions.

  • Graph D:

    • It has a vertical asymptote at x=2x = 2, but the function tends towards a positive value (crossing the x-axis again after x=2x = 2), which does not match the condition that the function approaches y=−4y = -4 as x→∞x \to \infty.

Conclusion:

The correct graph is Graph A.

Would you like more details on any of these steps, or do you have any questions?

Here are 5 related questions to further your understanding:

  1. How does a vertical asymptote affect the behavior of a function?
  2. What is the significance of horizontal asymptotes in graphing functions?
  3. Why does the function pass through the origin if f(0)=0f(0) = 0?
  4. What would change in the graph if lim⁡x→2+f(x)=∞\lim_{x \to 2^+} f(x) = \infty instead of −∞-\infty?
  5. How do the limits at x→±∞x \to \pm\infty help in determining the end behavior of the function?

Tip: When sketching functions, always start by analyzing critical points like where the function crosses the axes and its behavior at limits.

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Math Problem Analysis

Mathematical Concepts

Limits
Asymptotes
Graphing Functions
End Behavior

Formulas

lim x→2− f(x) = ∞
lim x→2+ f(x) = −∞
lim x→−∞ f(x) = −4
lim x→∞ f(x) = −4

Theorems

Limit Theorems
Asymptote Theorems

Suitable Grade Level

Grades 11-12 (Pre-Calculus/Calculus)

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