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Algebra 2 · F.BF.B.3

Transforming an exponential graph

The graph of y = 2^x hugs the x-axis. Where does the graph of y = 2^(x+3) - 5 level off, and which way does it run, without plotting it point by point?

You need
Graphing by transforming a parent graphAlgebra 2Graphs of exponential functionsAlgebra 1
Where you are 6 in Exponential functions
Functions13
Exponential functions6
Exponential growth and decay6Compound interest and compounding periodsGraphs of exponential functionsThe number e and the natural exponential function
Transforming an exponential graph
Writing an exponential function from a table or graph

When you reach for it

You know the graph of y = b^x from Graphs of exponential functions: it passes through (0, 1), runs flat along the x-axis on one side and grows without bound on the other. A rule such as y = 2^{x+3} - 5 is that graph moved, in exactly the sense of Graphing by transforming a parent graph. So instead of plotting it point by point, you move a few points of the parent and, above all, its asymptote.

You reach for this whenever an exponential rule carries extra numbers: a number added to the input, a number multiplying the power, a number added at the end. It also explains models that level off at something other than 0. A bowl of soup does not cool toward 0 degrees but toward the temperature of the room, so its graph is an exponential graph whose asymptote has been lifted.

The procedure

The general form. Every moved exponential in this lesson is written as in Combining transformations and the general form:

y = a \cdot b^{\,x - h} + k

with b > 0, b \ne 1 and a \ne 0. The parent is y = b^x. The number h shifts the graph right (h > 0) or left (h < 0), the number a stretches it vertically and flips it over if a < 0, and k shifts it up (k > 0) or down (k < 0). In y = 2^{x+3} - 5, the x + 3 is x - (-3), so h = -3, a = 1 and k = -5.

Carry a few points. The easiest points of the parent are \displaystyle \left(-1, \frac{1}{b}\right), (0, 1) and (1, b). Each point (p, q) of the parent lands at

(p,\ q) \;\mapsto\; (p + h,\ aq + k)

This is allowed because if q = b^p, then at the input x = p + h the new rule gives a \cdot b^{(p + h) - h} + k = a \cdot b^p + k = aq + k. So the moved point is really on the new graph, and every point of the new graph arises this way.

Carry the asymptote. The parent’s asymptote is the line y = 0, the line of all points of height 0. A height 0 goes to a \cdot 0 + k = k, so the new asymptote is the line y = k. Only k moves it. A horizontal shift slides the line y = 0 along itself, and a vertical stretch multiplies heights, and a times 0 is still 0. Draw the asymptote first, dashed; the curve is then laid against it.

Which side, and which end. The power b^{x-h} is positive for every x. If a > 0 then a \cdot b^{x-h} > 0, so y > k: the graph lies wholly above its asymptote, and the range is y > k. If a < 0 then a \cdot b^{x-h} < 0, so y < k: the graph lies wholly below, and the range is y < k. The domain is all real numbers, as for the parent. The base decides at which end the graph flattens onto the asymptote: on the left if b > 1 (as x \to -\infty, y \to k), on the right if 0 < b < 1. At the other end the graph runs off upward if a > 0 and downward if a < 0; this is the end behavior of the moved graph.

Intercepts. The y-intercept comes from putting x = 0 into the rule: y = a \cdot b^{-h} + k. It is the image of the parent’s point (0, 1) only when h = 0. An x-intercept exists only when 0 is in the range: when a > 0 and k < 0, or when a < 0 and k > 0. Then the graph crosses the x-axis exactly once, because it rises or falls steadily all the way. To find the crossing, solve a \cdot b^{x - h} + k = 0. When the power needed is a whole power of b the answer is exact; otherwise you can trap it between two integers, and the exact value needs a logarithm.

The gaps above the asymptote multiply by b. In y = a \cdot b^{x-h} + k, the part a \cdot b^{x-h} is y - k, the signed distance from the asymptote to the curve. Each step of 1 to the right multiplies that distance by b. The heights y themselves need not grow by a constant factor; the gaps y - k do. This is what you use to write a rule from a graph whose asymptote is not the x-axis.

What is special about exponentials. The Laws of exponents let three of the moves be rewritten as a change of a or b:

So every moved exponential can be rewritten as y = A \cdot B^x + k. The vertical shift is the one move that leaves the family y = a \cdot b^x, and it is exactly the move that shifts the asymptote.

Featurey = b^xy = a \cdot b^{\,x-h} + k
A key point(0, 1)(h,\ a + k)
The next point(1, b)(h + 1,\ ab + k)
Asymptotey = 0y = k
Rangey > 0y > k if a > 0; y < k if a < 0
Domainall real numbersall real numbers

Every row follows from the point rule and the asymptote rule above. The range can also be read off the finished drawing as a check.

Worked examples

Example 1

Shifted left and down

Graph y = 2^{x+3} - 5. Give its asymptote, range, y-intercept, and the two integers between which it crosses the x-axis.

Solution

Here b = 2, h = -3, a = 1 and k = -5: shift y = 2^x left 3 and down 5. Each point (p, q) goes to (p - 3,\ q - 5).

Parent pointImage
\displaystyle \left(-1, \frac{1}{2}\right)\displaystyle \left(-4, -\frac{9}{2}\right)
(0, 1)(-3, -4)
(1, 2)(-2, -3)
(2, 4)(-1, -1)
(3, 8)(0, 3)

The asymptote y = 0 moves down 5 to y = -5; the left shift does not touch it. Since a > 0 the graph lies above the asymptote, so the range is y > -5. The base 2 is above 1, so the graph flattens onto y = -5 on the left and climbs on the right.

−6 −4 2 −2 4 6 8 x y (−3, −4) (0, 3) y = −5
Figure 1Example 1. Every point of y = 2^x moves left 3 and down 5, and the asymptote moves down with it to y = -5.

The y-intercept: at x = 0, y = 2^3 - 5 = 3, the point (0, 3). It is the image of the parent’s (3, 8), not of (0, 1), because the graph was shifted sideways. The table shows y = -1 at x = -1 and y = 3 at x = 0. The graph is unbroken and rising, so it crosses the x-axis once, between x = -1 and x = 0. Check one point in the rule: at x = -2, 2^{1} - 5 = -3, as the table says.

Example 2

Flipped and lifted

Graph y = -3^{x} + 4. Give its asymptote, range and end behavior, and say between which two integers it crosses the x-axis.

Solution

The power is taken first, so -3^{x} means -\left(3^{x}\right), not (-3)^x. The rule is y = -1 \cdot 3^{x} + 4: a = -1, b = 3, h = 0, k = 4. Reflect y = 3^x in the x-axis, which makes every height negative, then shift up 4. Each point (p, q) goes to (p,\ 4 - q):

\begin{aligned} \left(-1, \tfrac{1}{3}\right) &\mapsto \left(-1, \tfrac{11}{3}\right) & (0, 1) &\mapsto (0, 3) \\ (1, 3) &\mapsto (1, 1) & (2, 9) &\mapsto (2, -5) \end{aligned}

The asymptote y = 0 goes to y = 4. Since a < 0 the graph lies below it, so the range is y < 4. The base 3 is above 1, so the flat end is on the left: as x \to -\infty, 3^x \to 0 and y \to 4, from below. On the right 3^x grows without bound, so as x \to \infty, y \to -\infty.

−4 −2 2 −4 −2 2 6 x y y = 4 (0, 3) (1, 1) (2, −5)
Figure 2Example 2. Flipped below its asymptote y = 4, the graph falls away on the right.

The graph is 1 at x = 1 and -5 at x = 2, and it falls steadily, so it crosses the x-axis once, between x = 1 and x = 2, where 3^x = 4. The y-intercept is (0, 3).

Example 3

A decaying base, stretched and shifted

Graph \displaystyle y = 4\left(\frac{1}{2}\right)^{x-1} + 1. Give its asymptote and range, and write the rule in the form y = A \cdot B^x + k.

Solution

Here \displaystyle b = \frac{1}{2}, h = 1, a = 4, k = 1: shift right 1, stretch vertically by 4, shift up 1. Each point (p, q) of \displaystyle y = \left(\frac{1}{2}\right)^x goes to (p + 1,\ 4q + 1):

\begin{aligned} (-1, 2) &\mapsto (0, 9) & (0, 1) &\mapsto (1, 5) \\ \left(1, \tfrac{1}{2}\right) &\mapsto (2, 3) & \left(2, \tfrac{1}{4}\right) &\mapsto (3, 2) \end{aligned}

The asymptote is y = 1. Since a > 0 the graph lies above it, and the range is y > 1. The base is between 0 and 1, so the graph falls and flattens onto y = 1 on the right. Notice the gaps above the asymptote: 8, 4, 2, 1, each half the one before, while the heights 9, 5, 3, 2 are not.

2 4 6 4 6 8 x y (0, 9) (1, 5) (2, 3) (3, 2) y = 1
Figure 3Example 3. A falling curve meets its asymptote on the right.

By Multiplying and dividing powers with the same base, and since \displaystyle \left(\frac{1}{2}\right)^{-1} = 2,

\begin{aligned} \left(\tfrac{1}{2}\right)^{x-1} &= \left(\tfrac{1}{2}\right)^{-1} \cdot \left(\tfrac{1}{2}\right)^x = 2\left(\tfrac{1}{2}\right)^x, \\ y &= 8\left(\tfrac{1}{2}\right)^x + 1. \end{aligned}

The shift right by 1 and the stretch by 4 together amount to a single stretch by 8. The y-intercept 8 + 1 = 9 agrees with the point (0, 9).

Example 4

Three rules, how many graphs?

Which of y = 3^{x+2}, y = 9 \cdot 3^x and y = 3^x + 9 have the same graph?

Solution

By Multiplying and dividing powers with the same base, 3^{x+2} = 3^x \cdot 3^2 = 9 \cdot 3^x for every x, so the first two rules give the same output at every input and have the same graph. Seen as moves, shifting y = 3^x left 2 and stretching it vertically by 9 give the same picture. For instance, the shift carries the parent’s (0, 1) to (-2, 1), and the stretch carries the parent’s \displaystyle \left(-2, \frac{1}{9}\right) up to (-2, 1) as well.

The third rule is different. Its y-intercept is 3^0 + 9 = 10, against 9 \cdot 3^0 = 9 for the other two, and its asymptote is y = 9, while theirs is still y = 0. Adding 9 after the power lifts the asymptote; adding 2 to the input, or multiplying by 9, never does.

Example 5

From words to a rule

Start from y = 2^x. Reflect it in the x-axis, stretch it vertically by 3, shift it right 1, then shift it up 6. Write the rule, and give the asymptote, range, y-intercept and x-intercept. What changes if the shift up 6 is done first?

Solution

Reflecting and stretching multiply every height by -3; shifting right 1 replaces x by x - 1; shifting up 6 adds 6 at the end. So

y = -3 \cdot 2^{x-1} + 6

The asymptote is y = 6. Since a = -3 < 0, the graph lies below it, and the range is y < 6. The y-intercept is \displaystyle -3 \cdot 2^{-1} + 6 = -\frac{3}{2} + 6 = \frac{9}{2}. For the x-intercept, set y = 0: 3 \cdot 2^{x-1} = 6, so 2^{x-1} = 2 = 2^1 and x - 1 = 1, giving x = 2. Check: -3 \cdot 2^{1} + 6 = 0.

The order of the vertical moves matters. If the shift up comes first, the reflection and stretch then multiply the whole height, 6 included: y = -3\left(2^{x-1} + 6\right) = -3 \cdot 2^{x-1} - 18. That graph has its asymptote at y = -18 and range y < -18, a different graph altogether.

Example 6

Reading the rule off a graph

The graph below is y = a \cdot b^x + k, with its asymptote drawn dashed at y = -2. Find a and b.

−4 2 2 6 8 x y (0, 1) (1, 4) (2, 10) y = −2
Figure 4The graph for Example 6.
Solution

The asymptote gives k = -2, so the rule is y = a \cdot b^x - 2. The point (0, 1) gives a \cdot b^0 - 2 = 1, and since b^0 = 1, a = 3. The point (1, 4) gives 3b - 2 = 4, so 3b = 6 and b = 2. The rule is

y = 3 \cdot 2^x - 2

Check with the third point: 3 \cdot 2^2 - 2 = 10, matching (2, 10). The heights 1, 4, 10 do not grow by a constant factor, but their gaps above the asymptote, 3, 6, 12, double at each step, which is the base 2. Working from \displaystyle \frac{4}{1} would have given the wrong base. The rule with k = 0 is Writing an exponential function from a table or graph; the asymptote is the one extra thing to read here.

Example 7

A bowl of soup cools

A bowl of soup at 80 degrees Celsius is left in a room at 20 degrees. Its temperature after t minutes is modeled by \displaystyle T = 60\left(\frac{1}{2}\right)^{t/10} + 20 (made-up numbers). Describe the graph as moves of \displaystyle T = \left(\frac{1}{2}\right)^t, find the temperature at t = 10, 20 and 30, give the asymptote and the range for t \ge 0, and say between which two of those times the soup first drops below 30 degrees.

Solution

Replacing t by \displaystyle \frac{t}{10} stretches the parent horizontally by a factor of 10: the parent’s point \displaystyle \left(1, \frac{1}{2}\right) moves to \displaystyle \left(10, \frac{1}{2}\right). Then the heights are stretched by 60 and lifted by 20. In words: the gap between the soup and the room starts at 80 - 20 = 60 degrees and halves every 10 minutes.

At t = 10: \displaystyle 60 \cdot \frac{1}{2} + 20 = 50. At t = 20: \displaystyle 60 \cdot \frac{1}{4} + 20 = 35. At t = 30: \displaystyle 60 \cdot \frac{1}{8} + 20 = 27.5.

10 20 30 40 50 60 20 40 60 80 t T T = 20
Figure 5Example 7. The soup cools toward the room’s temperature, 20 degrees, and never below it.

The asymptote is T = 20, the temperature of the room: the gap shrinks toward 0 but never reaches it, so the soup approaches room temperature without ever getting there. The domain is t \ge 0, the graph starts at the closed dot (0, 80) and falls, so the range is 20 < T \le 80. The temperature is 35 at t = 20 and 27.5 at t = 30, and it falls steadily, so it first drops below 30 degrees between 20 and 30 minutes.

Boundaries

  • The base needs b > 0 and b \ne 1, and the stretch needs a \ne 0. With a = 0 the rule is the horizontal line y = k, which has no asymptote and a range of one number.
  • A minus sign in front of a power reflects the graph: -3^x is -\left(3^x\right). A negative base, (-3)^x, is not an exponential function at all and has no unbroken graph.
  • The point rule reads a, h and k from the form a \cdot b^{\,x-h} + k exactly. A rule such as 3\left(2^x + 1\right) must be multiplied out first, to 3 \cdot 2^x + 3: its asymptote is y = 3, not y = 1. A rule such as 2^{3x - 6} must have its input factored, to 2^{3(x - 2)}, a horizontal compression by 3 and then a shift right 2.
  • The asymptote of a moved exponential is always horizontal, and only the vertical shift moves it. An exponential graph has no vertical asymptote.
  • In a context the graph is drawn only over a sensible domain, often t \ge 0, and the model’s asymptote is a value approached but never reached.

Where this goes next

The graph of a logarithm is the graph of an exponential reflected in the line y = x, and Graphs of logarithmic functions moves it by the same rules, with the roles swapped: there the asymptote is vertical, and the horizontal shift is the move that carries it. The exponential with base e in The number e and the natural exponential function moves exactly as here. Writing a rule from data, begun in Writing an exponential function from a table or graph and Building an exponential model from two points, reaches models with a lifted asymptote, such as cooling, through the gaps above the asymptote used in Example 6. And the line approached far out, here y = k, is the subject of Horizontal asymptotes for many more kinds of function.

Exercises

Graphing from the rule

  1. Describe how y = 2^{x-4} + 3 is moved from y = 2^x. Give its asymptote, range and y-intercept.
  2. Graph y = 3^{x+1} - 2 by moving the points \displaystyle \left(-1, \frac{1}{3}\right), (0, 1) and (1, 3) of y = 3^x. Give the asymptote, range and y-intercept.
  3. Graph \displaystyle y = -\left(\frac{1}{3}\right)^x - 2. Give the images of (-1, 3), (0, 1) and \displaystyle \left(1, \frac{1}{3}\right), the asymptote, the range and the end behavior.
  4. Graph y = 5 \cdot 2^x - 5. Give the images of the parent points at x = -1, 0 and 1, the asymptote and the range.
  5. Give the moves, the y-intercept, the point at x = 1, the asymptote and the range of \displaystyle y = \frac{1}{2} \cdot 4^x + 3.
  6. Describe y = 2 \cdot 3^{-x} + 1 as moves of y = 3^x. Does it rise or fall? Give its y-intercept, asymptote and range, and the points at x = -1 and x = 1.

Writing the rule

  1. Shift y = 3^x left 2 and down 4. Write the rule and give the y-intercept and asymptote.
  2. Reflect \displaystyle y = \left(\frac{1}{2}\right)^x in the x-axis, then shift it up 8. Write the rule, and give the y-intercept, the range and the x-intercept.
  3. Stretch y = 2^x vertically by 5, shift it right 3, then down 1. Write the rule and give the y-intercept and asymptote.
−2 2 x y (0, 5) (1, 7) (2, 11) y = 3 A x y (−1, −6) (0, −2) (1, 0) y = 2 B
Figure 6The graphs for exercise 10.
  1. Each graph in the exercise figure is y = a \cdot b^x + k, with its asymptote dashed. Write the rule for A (asymptote y = 3) and for B (asymptote y = 2).
  2. A graph of the form y = a \cdot b^x + k has asymptote y = 6 and passes through (0, 1) and (1, -9). Find the rule, and the height of the graph at x = -1.

One graph, two rules

  1. Show that y = 2^{x+5} and y = 32 \cdot 2^x have the same graph, and describe that graph as a move of y = 2^x in two ways.
  2. Show that stretching y = 4^x horizontally by a factor of 2, which gives y = 4^{x/2}, produces the graph of y = 2^x.
  3. Write y = 2^{-x+3} in the form y = A \cdot B^x.
  4. Which of y = 5^{x-1}, \displaystyle y = \frac{1}{5} \cdot 5^x and y = 5^x - 1 have the same graph? For the one that differs, give its y-intercept and asymptote.

Intercepts

  1. Find the x-intercept and y-intercept of y = 2^{x-1} - 8, and its asymptote.
  2. Does the graph of y = 4 \cdot 3^x + 1 cross the x-axis? Give its y-intercept and range.
  3. Find the x-intercept, y-intercept and range of y = -2 \cdot 3^x + 18.

In context

  1. A mug of cocoa cools in a room at 22 degrees Celsius, with temperature \displaystyle T = 50\left(\frac{1}{2}\right)^{t/12} + 22 after t minutes (made-up numbers). Find T at t = 0, 12, 24 and 36, give the asymptote and the range for t \ge 0, and say between which two of those times the cocoa first drops below 30 degrees.
  2. A learner’s typing speed after w weeks of practice is modeled by W = 70 - 40(0.8)^w words a minute (made-up numbers). Describe the graph as moves of W = (0.8)^w, find W at w = 0 and w = 3, and give the asymptote and the range for w \ge 0.

Reasoning

  1. Explain why shifting an exponential graph left or right never moves its asymptote, while shifting it up or down always does.
  2. Bartek says y = 2^x - 3 and y = 2^{x-3} are the same graph, since both “take away 3”. Compare their y-intercepts and asymptotes, and say how each is moved from y = 2^x.
  3. Explain why the graph of y = a \cdot b^x + k with a > 0 never meets the line y = k, and why, when also k < 0, it crosses the x-axis exactly once.
  4. Show that y = a \cdot b^{c(x-h)} + k, with b > 0, b \ne 1 and c \ne 0, can always be rewritten as y = A \cdot B^x + k, and say what A and B are.
Answers
  1. Right 4 and up 3. Asymptote y = 3; range y > 3; y-intercept \displaystyle 2^{-4} + 3 = \frac{49}{16}, the point \displaystyle \left(0, \frac{49}{16}\right).
  2. Each point moves to (p - 1,\ q - 2): \displaystyle \left(-2, -\frac{5}{3}\right), (-1, -1) and (0, 1). Asymptote y = -2; range y > -2; y-intercept (0, 1), since 3^1 - 2 = 1.
  3. Each point goes to (p,\ -q - 2): (-1, -5), (0, -3), \displaystyle \left(1, -\frac{7}{3}\right). Asymptote y = -2, on the right since the base is below 1; range y < -2. As x \to \infty, y \to -2; as x \to -\infty, y \to -\infty.
  4. Each point goes to (p,\ 5q - 5): \displaystyle \left(-1, -\frac{5}{2}\right), (0, 0), (1, 5). The graph passes through the origin. Asymptote y = -5; range y > -5.
  5. Compress vertically by \displaystyle \frac{1}{2}, then shift up 3. y-intercept \displaystyle \frac{1}{2} + 3 = \frac{7}{2}; at x = 1, \displaystyle \frac{1}{2} \cdot 4 + 3 = 5. Asymptote y = 3; range y > 3.
  6. Reflect in the y-axis, stretch vertically by 2, shift up 1; it is \displaystyle y = 2\left(\frac{1}{3}\right)^x + 1, so it falls. y-intercept (0, 3); asymptote y = 1, on the right; range y > 1. Points (-1, 7) and \displaystyle \left(1, \frac{5}{3}\right).
  7. y = 3^{x+2} - 4. y-intercept 3^2 - 4 = 5; asymptote y = -4.
  8. \displaystyle y = -\left(\frac{1}{2}\right)^x + 8. y-intercept -1 + 8 = 7; range y < 8. Setting y = 0: \displaystyle \left(\frac{1}{2}\right)^x = 8 = \left(\frac{1}{2}\right)^{-3}, so the x-intercept is x = -3.
  9. y = 5 \cdot 2^{x-3} - 1. y-intercept \displaystyle \frac{5}{8} - 1 = -\frac{3}{8}; asymptote y = -1.
  10. A: k = 3; (0, 5) gives a = 2; (1, 7) gives 2b + 3 = 7, b = 2. So y = 2 \cdot 2^x + 3; check (2, 11): 8 + 3 = 11. B: k = 2; (0, -2) gives a = -4; (1, 0) gives -4b + 2 = 0, \displaystyle b = \frac{1}{2}. So \displaystyle y = -4\left(\frac{1}{2}\right)^x + 2; check (-1, -6): -8 + 2 = -6.
  11. a + 6 = 1 gives a = -5; -5b + 6 = -9 gives b = 3. So y = -5 \cdot 3^x + 6, and at x = -1 the height is \displaystyle -\frac{5}{3} + 6 = \frac{13}{3}.
  12. By product of powers, 2^{x+5} = 2^5 \cdot 2^x = 32 \cdot 2^x for every x. The graph is y = 2^x shifted left 5, or y = 2^x stretched vertically by 32; its y-intercept is 32.
  13. By power of a power, 4^{x/2} = \left(4^{1/2}\right)^x = 2^x for every x, so the two rules have the same graph.
  14. \displaystyle 2^{-x+3} = 2^3 \cdot 2^{-x} = 8\left(\frac{1}{2}\right)^x, so A = 8 and \displaystyle B = \frac{1}{2}.
  15. The first two: \displaystyle 5^{x-1} = 5^{-1} \cdot 5^x = \frac{1}{5} \cdot 5^x. The third, y = 5^x - 1, has y-intercept 0 and asymptote y = -1 (the others have \displaystyle \frac{1}{5} and y = 0).
  16. 2^{x-1} = 8 = 2^3, so x - 1 = 3 and the x-intercept is x = 4. y-intercept \displaystyle \frac{1}{2} - 8 = -\frac{15}{2}; asymptote y = -8.
  17. No. Since 4 \cdot 3^x > 0, every output is above 1: the range is y > 1, which does not include 0. y-intercept 4 + 1 = 5.
  18. 2 \cdot 3^x = 18 gives 3^x = 9, so the x-intercept is x = 2. y-intercept -2 + 18 = 16; range y < 18.
  19. 72, 47, 34.5 and 28.25 degrees. Asymptote T = 22, the room’s temperature; range 22 < T \le 72. It is 34.5 at t = 24 and 28.25 at t = 36, so it first drops below 30 between 24 and 36 minutes.
  20. W = -40(0.8)^w + 70: stretch vertically by 40, reflect in the w-axis, shift up 70. At w = 0, W = 30; at w = 3, W = 70 - 40 \cdot 0.512 = 49.52. Asymptote W = 70; range 30 \le W < 70.
  21. The asymptote is the horizontal line y = 0. A left or right shift slides every point sideways, and a horizontal line slid sideways is the same line, so the asymptote stays put. An up or down shift by k changes every height by k, the asymptote’s height 0 included, so it becomes y = k.
  22. They differ. y = 2^x - 3 is y = 2^x shifted down 3: y-intercept 1 - 3 = -2, asymptote y = -3. y = 2^{x-3} is y = 2^x shifted right 3: y-intercept \displaystyle 2^{-3} = \frac{1}{8}, asymptote y = 0.
  23. With a > 0 and b > 0, a \cdot b^x is positive for every x, so y = a \cdot b^x + k is always greater than k: it never equals k. Its outputs are all the numbers above k, since a \cdot b^x takes every positive value, so when k < 0 the height 0 is among them. The graph rises or falls steadily, so it reaches each height once and crosses the x-axis exactly once.
  24. By the laws of exponents, b^{c(x-h)} = b^{cx - ch} = b^{-ch} \cdot \left(b^c\right)^x. So y = \left(a \cdot b^{-ch}\right) \cdot \left(b^c\right)^x + k, with A = a \cdot b^{-ch} and B = b^c, which is positive and not 1 because c \ne 0.
kg1128 · updated 2026-10-19 · CC-BY-SA-4.0