Math Problem Statement

how to graph y=e^(2x)-3e^x+2

Solution

To graph the function y=e2x3ex+2y = e^{2x} - 3e^x + 2, follow these steps:

Step 1: Understand the behavior of the function.

The function consists of exponential terms. Let's break it down:

  • e2xe^{2x}: This grows rapidly as xx increases, since the exponent is 2x2x.
  • 3ex-3e^x: This term decays less rapidly than e2xe^{2x}, as the exponent is linear in xx.
  • +2+2: This is a constant shift upward by 2 units.

Step 2: Find the intercepts.

  • Y-intercept (when x=0x = 0): y(0)=e2(0)3e0+2=e03e0+2=13+2=0y(0) = e^{2(0)} - 3e^{0} + 2 = e^0 - 3e^0 + 2 = 1 - 3 + 2 = 0 So, the y-intercept is at (0,0)(0, 0).

  • X-intercepts (when y=0y = 0): To find the x-intercepts, solve: e2x3ex+2=0e^{2x} - 3e^x + 2 = 0 Let u=exu = e^x, so e2x=u2e^{2x} = u^2. The equation becomes: u23u+2=0u^2 - 3u + 2 = 0 Factor the quadratic: (u1)(u2)=0(u - 1)(u - 2) = 0 So, u=1u = 1 or u=2u = 2. Therefore:

    • If u=1u = 1, ex=1e^x = 1, so x=0x = 0.
    • If u=2u = 2, ex=2e^x = 2, so x=ln(2)0.693x = \ln(2) \approx 0.693.

Thus, the x-intercepts are at x=0x = 0 and x0.693x \approx 0.693.

Step 3: Find the asymptotic behavior.

  • As xx \to \infty, e2xe^{2x} grows much faster than exe^x or the constant 2. Therefore, the function will grow without bound as xx \to \infty.
  • As xx \to -\infty, both e2xe^{2x} and exe^x approach 0, so the function approaches 2 (the constant term).

Thus, the horizontal asymptote is y=2y = 2.

Step 4: Plot key points and asymptotes.

  • Plot the intercepts (0,0)(0, 0) and (ln(2),0)(\ln(2), 0).
  • Plot some additional points for negative and positive values of xx (e.g., for x=1,1,2x = -1, 1, 2).
  • Draw the horizontal asymptote at y=2y = 2.

Step 5: Sketch the graph.

  • The graph should start near the horizontal asymptote of y=2y = 2 as xx \to -\infty.
  • It passes through the intercepts, with the function rising steeply as xx increases, due to the e2xe^{2x} term dominating for large positive xx.

Summary:

  • Intercepts: (0,0)(0, 0) and (ln(2),0)(\ln(2), 0).
  • Horizontal asymptote: y=2y = 2.
  • End behavior: As xx \to -\infty, y2y \to 2, and as xx \to \infty, yy \to \infty.

Would you like to explore more about this function, or have any further questions?

Here are some related questions:

  1. How do exponential functions behave for different values of the base (e.g., axa^x vs exe^x)?
  2. What are the differences in the graphs of y=exy = e^x and y=e2xy = e^{2x}?
  3. How can you use derivatives to analyze the shape of the graph?
  4. What are the applications of the function y=e2x3ex+2y = e^{2x} - 3e^x + 2?
  5. How does the horizontal asymptote relate to the function's behavior at large values of xx?

Tip: To graph any exponential function accurately, always consider its asymptotes and behavior at extreme values of xx.

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Math Problem Analysis

Mathematical Concepts

Exponential Functions
Graphing Functions
Intercepts
Asymptotes

Formulas

y = e^(2x) - 3e^x + 2
e^x = 1 or e^x = 2

Theorems

Behavior of exponential functions
Intercepts of functions
Horizontal asymptotes

Suitable Grade Level

Grades 11-12