Plotting a Growth Curve from a Table

Graph $y = 3 \cdot 2^x$ for $x = -2$ to $x = 3$. State the y-intercept, the asymptote, the domain and the range.

Compute the values for negative $x$: $3 \cdot 2^{-2} = 3 \cdot \frac{1}{4} = 0.75$ and $3 \cdot 2^{-1} = 3 \cdot \frac{1}{2} = 1.5$ = Points $(-2, 0.75)$ and $(-1, 1.5)$

Compute the values for $x = 0$ to $3$: $3 \cdot 2^0 = 3$, $3 \cdot 2^1 = 6$, $3 \cdot 2^2 = 12$, $3 \cdot 2^3 = 24$ = Points $(0, 3)$, $(1, 6)$, $(2, 12)$, $(3, 24)$

Plot and join the points: Each step to the right doubles $y$; each step to the left halves it = A rising curve, flat on the left and steep on the right

Read the key features: $a = 3$ gives the y-intercept; nothing is added, so $k = 0$ = y-intercept $(0, 3)$, asymptote $y = 0$, domain all real numbers, range $y > 0$

Answer: The graph passes through $(-2, 0.75)$, $(-1, 1.5)$, $(0, 3)$, $(1, 6)$, $(2, 12)$, $(3, 24)$. It rises from left to right, crosses the y-axis at $(0, 3)$, approaches the asymptote $y = 0$ on the left, has domain all real numbers and range $y > 0$.

A Falling Curve: Base Between 0 and 1

Graph $y = 8 \cdot (0.5)^x$ for $x = -2$ to $x = 3$ and compare it with the graph of $y = 8 \cdot 2^x$.

Compute the values: $8 \cdot (0.5)^{-2} = 8 \cdot 4 = 32$, $8 \cdot (0.5)^{-1} = 16$, $8 \cdot (0.5)^0 = 8$, $8 \cdot (0.5)^1 = 4$, $8 \cdot (0.5)^2 = 2$, $8 \cdot (0.5)^3 = 1$ = Points $(-2, 32)$, $(-1, 16)$, $(0, 8)$, $(1, 4)$, $(2, 2)$, $(3, 1)$

Describe the shape: Each step to the right halves $y$, so the curve falls and flattens out toward the x-axis on the right = Falling curve, asymptote $y = 0$

Compare with $y = 8 \cdot 2^x$: $(0.5)^x = (2^{-1})^x = 2^{-x}$, so $8 \cdot (0.5)^x = 8 \cdot 2^{-x}$: the value at $x$ equals the value of $8 \cdot 2^x$ at $-x$ = The two graphs are mirror images in the y-axis

Answer: The graph falls from $(-2, 32)$ through $(0, 8)$ to $(3, 1)$ and approaches $y = 0$ on the right. It is the mirror image of $y = 8 \cdot 2^x$ in the y-axis; both share the y-intercept $(0, 8)$, the asymptote $y = 0$ and the range $y > 0$.

Shifting a Graph Right and Up

Sketch $y = 2^{x-1} + 3$. Find the asymptote, the y-intercept, the domain and the range.

Identify the transformation: Compare with $y = a \cdot b^{x-h} + k$: $a = 1$, $b = 2$, $h = 1$, $k = 3$ = Shift $y = 2^x$ right 1 and up 3

Move three points of $y = 2^x$: $(0, 1) \to (1, 4)$, $(1, 2) \to (2, 5)$, $(2, 4) \to (3, 7)$: add 1 to each $x$ and 3 to each $y$ = Points $(1, 4)$, $(2, 5)$, $(3, 7)$

Move the asymptote: The asymptote $y = 0$ moves up 3 units; the horizontal shift does not change it = Asymptote $y = 3$

Find the y-intercept: Put $x = 0$: $2^{0-1} + 3 = 0.5 + 3 = 3.5$ = y-intercept $(0, 3.5)$

State domain and range: $a > 0$, so the graph stays above $y = 3$ = Domain all real numbers, range $y > 3$

Answer: The graph is $y = 2^x$ moved 1 right and 3 up: it passes through $(0, 3.5)$, $(1, 4)$, $(2, 5)$ and $(3, 7)$, has the asymptote $y = 3$, domain all real numbers and range $y > 3$.

A Reflected Graph Below Its Asymptote

Make a table for $y = -2^x + 4$ from $x = -2$ to $x = 3$. Describe the graph and find where it crosses the x-axis.

Read the expression correctly: $-2^x$ means $-(2^x)$: compute $2^x$ first, then change the sign. So $a = -1$, $b = 2$, $k = 4$ = Reflection of $y = 2^x$ in the x-axis, then up 4

Compute the values: $x = -2$: $-0.25 + 4 = 3.75$; $x = -1$: $-0.5 + 4 = 3.5$; $x = 0$: $-1 + 4 = 3$; $x = 1$: $-2 + 4 = 2$; $x = 2$: $-4 + 4 = 0$; $x = 3$: $-8 + 4 = -4$ = Points $(-2, 3.75)$, $(-1, 3.5)$, $(0, 3)$, $(1, 2)$, $(2, 0)$, $(3, -4)$

Find the asymptote and range: As $x$ goes far to the left, $2^x$ gets close to 0, so $y$ gets close to 4 from below = Asymptote $y = 4$, range $y < 4$

Find the x-intercept: $-2^x + 4 = 0 \Rightarrow 2^x = 4 = 2^2 \Rightarrow x = 2$ = x-intercept $(2, 0)$

Answer: The graph falls from just below $y = 4$ on the left, through $(0, 3)$ and $(2, 0)$, and drops steeply to $(3, -4)$. Asymptote $y = 4$, domain all real numbers, range $y < 4$, x-intercept $(2, 0)$, y-intercept $(0, 3)$.

Finding the Equation from Two Points

The graph of $y = a \cdot b^x$ passes through $(1, 6)$ and $(3, 24)$. Find $a$ and $b$, then find the y-intercept.

Write an equation for each point: $a \cdot b^1 = 6$ and $a \cdot b^3 = 24$ = Two equations with $a$ and $b$

Divide the second equation by the first: $\frac{a \cdot b^3}{a \cdot b} = \frac{24}{6} \Rightarrow b^2 = 4$ = $b = 2$ (the base must be positive)

Substitute $b$ to find $a$: $a \cdot 2 = 6 \Rightarrow a = 3$ = $a = 3$

Write the equation and check: $y = 3 \cdot 2^x$; check $(3, 24)$: $3 \cdot 2^3 = 3 \cdot 8 = 24$ = y-intercept $(0, 3)$

Answer: $y = 3 \cdot 2^x$, so $a = 3$, $b = 2$ and the y-intercept is $(0, 3)$. With the points $(0, 3)$ and $(1, 6)$ it is even quicker: $a$ is the y-value at $x = 0$, and $b = 6 \div 3 = 2$.

Mistake: Keeping the asymptote at $y = 0$ after a vertical shift, for example for $y = 2^x + 3$

Why: Adding $k$ moves every point up by $k$, including the level the curve approaches. The graph of $y = 2^x + 3$ never goes below 3.

Correct: For $y = a \cdot b^{x-h} + k$ the asymptote is $y = k$. For $y = 2^x + 3$ it is $y = 3$ and the range is $y > 3$.

Mistake: Shifting the wrong way: thinking $y = 2^{x-3}$ moves the graph 3 units left

Why: In $x - 3$ you need a larger $x$ to get the same exponent as before. The point $(0, 1)$ of $y = 2^x$ is now at $(3, 1)$.

Correct: $x - h$ shifts right by $h$, $x + h$ shifts left by $h$. $y = 2^{x-3}$ moves right 3, $y = 2^{x+3}$ moves left 3.

Mistake: Reading the y-intercept as $a$ (or $a + k$) when the graph is also shifted sideways

Why: The y-intercept is $(0, a)$ only for $y = a \cdot b^x$. After a horizontal shift the exponent at $x = 0$ is no longer 0.

Correct: Always substitute $x = 0$. For $y = 2^{x-1} + 3$: $2^{-1} + 3 = 3.5$, not 1 and not 4.

Mistake: Treating $-2^x$ as $(-2)^x$

Why: Powers are computed before the minus sign, so $-2^x = -(2^x)$. The base is 2, not $-2$, and $(-2)^x$ is not even an exponential function.

Correct: $-2^x$ is the graph of $2^x$ reflected in the x-axis: at $x = 2$ it gives $-4$, while $(-2)^2 = 4$.

Reading Lab Data on a Graph

Biologists plot cell counts over time. If the points lie on an exponential curve, two readings are enough to write the model and predict later counts.

A culture counted 100 cells at the start and 200 cells one hour later follows $y = 100 \cdot 2^x$, so after 4 hours the curve is at $100 \cdot 16 = 1600$ cells.

Video Views with a Baseline

Some quantities have a part that grows exponentially on top of a constant part. The constant part is the vertical shift $k$, and on the graph it is the asymptote.

A video gets a steady 100 views a day from search, plus a shared part that doubles every day: $V(t) = 25 \cdot 2^t + 100$. Going back in time the shared part shrinks toward 0, so the graph approaches $V = 100$.

Folding Paper

Each fold doubles the thickness of a sheet of paper, so the thickness after $n$ folds is an exponential function whose graph climbs faster and faster.

A sheet 0.1 mm thick has thickness $T(n) = 0.1 \cdot 2^n$ mm. After 7 folds that is $0.1 \cdot 128 = 12.8$ mm.

Q: Does an exponential graph ever touch its asymptote?

A: No. For $y = a \cdot b^x + k$, the term $a \cdot b^x$ is never 0, so $y$ is never equal to $k$. The curve gets as close as you like, which is why the range is $y > k$ or $y < k$ and never includes $k$.

Q: How many points do I need to sketch the graph?

A: Four or five are enough: the y-intercept, one or two points on each side, and the asymptote. A table from $x = -2$ to $x = 3$ shows both the flat side and the steep side.

Q: Why do $y = 2^x$ and $y = (0.5)^x$ look like mirror images?

A: Because $(0.5)^x = 2^{-x}$. The value of $(0.5)^x$ at $x = 3$ equals the value of $2^x$ at $x = -3$, so every point $(x, y)$ of one graph has the partner $(-x, y)$ on the other.

Q: Does a horizontal shift change the asymptote?

A: No. Shifting left or right slides the curve along the asymptote, so it stays at $y = k$. Only the vertical shift $k$ moves the asymptote. A horizontal shift does change the y-intercept, though.