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Functions - Exponents – exponential function

Grade 11IB_AA

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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An exponential function is defined by the form f(x)=a⋅bx+kf(x) = a \cdot b^x + k, where b>0b > 0 and b≠1b \neq 1. The base bb determines the growth (b>1b > 1) or decay (0<b<10 < b < 1) of the function, while kk represents the horizontal asymptote y=ky = k.

Graph of an exponential growth function y = 2^x showing the horizontal asymptote at y = 0.
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The natural exponential function uses the base e≈2.718e \approx 2.718. This base is fundamental in modeling continuous growth and decay, such as compound interest and radioactive decay, typically written as f(x)=exf(x) = e^x.

Graph of the natural exponential function y = e^x passing through (0,1).
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Exponential decay occurs when the base bb is between 00 and 11. As xx increases, the value of f(x)f(x) approaches the horizontal asymptote from above.

Graph of an exponential decay function y = (1/2)^x.
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Vertical translations shift the entire graph and its asymptote. For f(x)=bx+kf(x) = b^x + k, the horizontal asymptote is at y=ky = k and the yy-intercept is at (0,1+k)(0, 1+k).

📐Formulae

f(x)=a⋅bx+kf(x) = a \cdot b^{x} + k

f(x)=exf(x) = e^{x}

A=P(1+r100n)ntA = P\left(1 + \frac{r}{100n}\right)^{nt}

A=PertA = Pe^{rt}

y=k (Horizontal Asymptote)y = k \text{ (Horizontal Asymptote)}

💡Examples

Problem 1:

Given the function f(x)=5⋅2x−10f(x) = 5 \cdot 2^{x} - 10, find the horizontal asymptote, the yy-intercept, and the xx-intercept.

Solution:

  1. Horizontal Asymptote: The constant term added to the exponential part is −10-10. Thus, the horizontal asymptote is y=−10y = -10.
  2. yy-intercept: Set x=0x = 0. f(0)=5⋅20−10=5(1)−10=−5f(0) = 5 \cdot 2^{0} - 10 = 5(1) - 10 = -5 The yy-intercept is (0,−5)(0, -5).
  3. xx-intercept: Set f(x)=0f(x) = 0. 5⋅2x−10=05 \cdot 2^{x} - 10 = 0 5⋅2x=105 \cdot 2^{x} = 10 2x=22^{x} = 2 x=1x = 1 The xx-intercept is (1,0)(1, 0).

Explanation:

To find intercepts, we alternate setting xx and yy to zero. The horizontal asymptote is determined by the vertical shift of the parent exponential function.

Problem 2:

A population of bacteria grows according to the model P(t)=1000e0.05tP(t) = 1000e^{0.05t}, where tt is time in hours. Find the initial population and the population after 10 hours.

Solution:

  1. Initial population: Set t=0t = 0. P(0)=1000e0.05(0)=1000e0=1000⋅1=1000P(0) = 1000e^{0.05(0)} = 1000e^{0} = 1000 \cdot 1 = 1000
  2. Population after 10 hours: P(10)=1000e0.05(10)=1000e0.5P(10) = 1000e^{0.05(10)} = 1000e^{0.5} Using a calculator: P(10)≈1000⋅1.6487=1648.7P(10) \approx 1000 \cdot 1.6487 = 1648.7 Population ≈1649\approx 1649 bacteria.

Explanation:

The coefficient 10001000 represents the initial value P0P_0. The exponent 0.050.05 represents the continuous growth rate of 5%5\%.

Problem 3:

Solve for xx: 9x−1=(13)x+29^{x-1} = \left(\frac{1}{3}\right)^{x+2}

Solution:

Write both sides with the same base 33: (32)x−1=(3−1)x+2(3^{2})^{x-1} = (3^{-1})^{x+2} 32x−2=3−x−23^{2x-2} = 3^{-x-2} Equate the exponents: 2x−2=−x−22x - 2 = -x - 2 3x=03x = 0 x=0x = 0

Explanation:

When solving exponential equations without logarithms, try to express both sides as powers of the same base and then set the exponents equal to each other.

Problem 4:

Sketch the graph of f(x)=2x−1+3f(x) = 2^{x-1} + 3 and state the equation of its horizontal asymptote and the yy-intercept.

Graph of y = 2^(x-1) + 3 with asymptote at y = 3 and y-intercept at 3.5.

Solution:

  1. Asymptote: The constant term is +3+3, so the horizontal asymptote is y=3y = 3.
  2. y-intercept: Set x=0x = 0: f(0)=20−1+3=2−1+3=0.5+3=3.5f(0) = 2^{0-1} + 3 = 2^{-1} + 3 = 0.5 + 3 = 3.5 The yy-intercept is (0,3.5)(0, 3.5).
  3. Horizontal Shift: The term (x−1)(x-1) shifts the graph of 2x2^x one unit to the right.

Explanation:

To graph an exponential function, first identify the horizontal asymptote, then calculate the yy-intercept and one or two additional points to determine the shape.

Problem 5:

Determine the value of kk for the function f(x)=3⋅ekxf(x) = 3 \cdot e^{kx} if the graph passes through the point (2,12)(2, 12).

Graph of an exponential function with base e passing through (2, 12) and (0, 3).

Solution:

  1. Substitute the point (2,12)(2, 12) into the equation: 12=3⋅e2k12 = 3 \cdot e^{2k}
  2. Divide by 33: 4=e2k4 = e^{2k}
  3. Take the natural logarithm of both sides: ln⁡(4)=ln⁡(e2k)\ln(4) = \ln(e^{2k}) ln⁡(4)=2k\ln(4) = 2k
  4. Solve for kk: k=ln⁡(4)2=ln⁡(22)2=2ln⁡(2)2=ln⁡(2)≈0.693k = \frac{\ln(4)}{2} = \frac{\ln(2^2)}{2} = \frac{2 \ln(2)}{2} = \ln(2) \approx 0.693

Explanation:

When given a point on an exponential curve, substitute the coordinates to solve for the unknown parameter using logarithms.

Exponents – exponential function Grade 11 Notes & Examples