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Plot the following exponential functions, identifying the horizontal asymptote, y-intercept, and domain/range.

Mathematics

ScanSolve AI Answer

Graph of f(x) = −(1/2)^x + 4

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Key points of f(x) = −(1/2)^x + 4
Pointxy
x-intercept−20
y-intercept03

Step-by-step solution

12 steps

To plot these exponential functions, we'll identify key features like the horizontal asymptote, y-intercept, and a few points.

Part a· 4 steps

f(x)=−2−x+4f(x) = -2^{-x} + 4

This function can be rewritten as f(x)=−(12)x+4f(x) = -\left(\frac{1}{2}\right)^x + 4.

  1. Identify the horizontal asymptote.

    The term +4+4 indicates a vertical shift upwards by 4 units.
    The horizontal asymptote is at y=4y = \boxed{4}.

  2. Find the y-intercept.

    Set x=0x=0:

    f(0)=−2−0+4=−1+4=3f(0) = -2^{-0} + 4 = -1 + 4 = 3

    The y-intercept is (0,3)\boxed{(0, 3)}.

  3. Calculate additional points for plotting.

    xxf(x)=−2−x+4f(x) = -2^{-x} + 4Point
    -2−2−(−2)+4=−22+4=−4+4=0-2^{-(-2)} + 4 = -2^2 + 4 = -4 + 4 = 0(−2,0)(-2, 0)
    -1−2−(−1)+4=−21+4=−2+4=2-2^{-(-1)} + 4 = -2^1 + 4 = -2 + 4 = 2(−1,2)(-1, 2)
    0−20+4=−1+4=3-2^0 + 4 = -1 + 4 = 3(0,3)(0, 3)
    1−2−1+4=−12+4=3.5-2^{-1} + 4 = -\frac{1}{2} + 4 = 3.5(1,3.5)(1, 3.5)
    2−2−2+4=−14+4=3.75-2^{-2} + 4 = -\frac{1}{4} + 4 = 3.75(2,3.75)(2, 3.75)
  4. State the domain and range.

    The domain is all real numbers: x∈(−∞,∞)\boxed{x \in (-\infty, \infty)}.
    The range is all real numbers less than the horizontal asymptote: y∈(−∞,4)\boxed{y \in (-\infty, 4)}.

Part b· 4 steps

f(x)=−2(2)x+2f(x) = -2(2)^x + 2

  1. Identify the horizontal asymptote.

    The term +2+2 indicates a vertical shift upwards by 2 units.
    The horizontal asymptote is at y=2y = \boxed{2}.

  2. Find the y-intercept.

    Set x=0x=0:

    f(0)=−2(2)0+2=−2(1)+2=−2+2=0f(0) = -2(2)^0 + 2 = -2(1) + 2 = -2 + 2 = 0

    The y-intercept is (0,0)\boxed{(0, 0)}.

  3. Calculate additional points for plotting.

    xxf(x)=−2(2)x+2f(x) = -2(2)^x + 2Point
    -2−2(2)−2+2=−2(14)+2=−12+2=1.5-2(2)^{-2} + 2 = -2(\frac{1}{4}) + 2 = -\frac{1}{2} + 2 = 1.5(−2,1.5)(-2, 1.5)
    -1−2(2)−1+2=−2(12)+2=−1+2=1-2(2)^{-1} + 2 = -2(\frac{1}{2}) + 2 = -1 + 2 = 1(−1,1)(-1, 1)
    0−2(2)0+2=−2(1)+2=0-2(2)^0 + 2 = -2(1) + 2 = 0(0,0)(0, 0)
    1−2(2)1+2=−4+2=−2-2(2)^1 + 2 = -4 + 2 = -2(1,−2)(1, -2)
    2−2(2)2+2=−2(4)+2=−8+2=−6-2(2)^2 + 2 = -2(4) + 2 = -8 + 2 = -6(2,−6)(2, -6)
  4. State the domain and range.

    The domain is all real numbers: x∈(−∞,∞)\boxed{x \in (-\infty, \infty)}.
    The range is all real numbers less than the horizontal asymptote: y∈(−∞,2)\boxed{y \in (-\infty, 2)}.

Part c· 4 steps

f(x)=3(12)x+2f(x) = 3\left(\frac{1}{2}\right)^x + 2

  1. Identify the horizontal asymptote.

    The term +2+2 indicates a vertical shift upwards by 2 units.
    The horizontal asymptote is at y=2y = \boxed{2}.

  2. Find the y-intercept.

    Set x=0x=0:

    f(0)=3(12)0+2=3(1)+2=3+2=5f(0) = 3\left(\frac{1}{2}\right)^0 + 2 = 3(1) + 2 = 3 + 2 = 5

    The y-intercept is (0,5)\boxed{(0, 5)}.

  3. Calculate additional points for plotting.

    xxf(x)=3(12)x+2f(x) = 3\left(\frac{1}{2}\right)^x + 2Point
    -23(12)−2+2=3(4)+2=12+2=143\left(\frac{1}{2}\right)^{-2} + 2 = 3(4) + 2 = 12 + 2 = 14(−2,14)(-2, 14)
    -13(12)−1+2=3(2)+2=6+2=83\left(\frac{1}{2}\right)^{-1} + 2 = 3(2) + 2 = 6 + 2 = 8(−1,8)(-1, 8)
    03(12)0+2=3(1)+2=53\left(\frac{1}{2}\right)^0 + 2 = 3(1) + 2 = 5(0,5)(0, 5)
    13(12)1+2=3(12)+2=1.5+2=3.53\left(\frac{1}{2}\right)^1 + 2 = 3(\frac{1}{2}) + 2 = 1.5 + 2 = 3.5(1,3.5)(1, 3.5)
    23(12)2+2=3(14)+2=0.75+2=2.753\left(\frac{1}{2}\right)^2 + 2 = 3(\frac{1}{4}) + 2 = 0.75 + 2 = 2.75(2,2.75)(2, 2.75)
  4. State the domain and range.

    The domain is all real numbers: x∈(−∞,∞)\boxed{x \in (-\infty, \infty)}.
    The range is all real numbers greater than the horizontal asymptote: y∈(2,∞)\boxed{y \in (2, \infty)}.

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