Unit 7 Homework 5 Graphing Logarithmic Functions Answers

Introduction

Understanding how to graph logarithmic functions is a fundamental skill in Algebra II and Pre‑Calculus, and many students encounter this topic in Unit 7, Homework 5 of their curriculum. This article walks you through the complete solution process, from identifying the function’s key features to plotting accurate graphs. By the end, you’ll not only have the answers for the assigned problems but also a solid conceptual framework that lets you tackle any similar task with confidence.

Why Logarithmic Graphs Matter

Logarithmic functions model real‑world phenomena such as pH levels, earthquake intensity, and population growth under constraints. Mastering their graphs helps you:

• Visualize how changes in the base affect the curve.
• Translate equations into meaningful data trends.
• Prepare for standardized tests that test both procedural fluency and conceptual insight.

Core Concepts for Graphing Logarithmic Functions

1. General Form and Base

A logarithmic function is written as

[ f(x)=\log_b (x-h)+k ]

where b > 0, b ≠ 1 is the base, h shifts the graph horizontally, and k moves it vertically That's the part that actually makes a difference..

• If b > 1, the graph rises to the right.
• If 0 < b < 1, the graph falls to the right.

2. Domain and Range

• Domain: (x>h) (the argument of the log must be positive).
• Range: All real numbers, ((-\infty,\infty)), because the output can be any real number after vertical shifts.

3. Asymptote

The vertical line (x = h) is a vertical asymptote; the graph never crosses it but approaches it infinitely.

4. Intercepts

• x‑intercept: Set (f(x)=0) → (\log_b (x-h) = -k) → (x = h + b^{-k}).
• y‑intercept: Exists only if (h < 0); plug (x=0) into the function.

5. Reference Points for Plotting

Because logarithms grow slowly, it’s useful to compute values at:

• (x = h + 1) (gives (f = \log_b 1 + k = k)).
• (x = h + b) (gives (f = \log_b b + k = 1 + k)).
• (x = h + b^2) (gives (f = 2 + k)).

These points create a “ladder” that guides the shape of the curve Practical, not theoretical..

Step‑by‑Step Solution for Homework 5

Below is a typical set of problems you might find in Unit 7, Homework 5. The solutions illustrate the systematic approach you should adopt for every logarithmic graph.

Problem 1: Graph (f(x)=\log_2 (x-3) + 1)

1. Identify parameters:

   • Base (b = 2) (greater than 1, so the graph rises).
   • Horizontal shift (h = 3).
   • Vertical shift (k = 1).

2. Domain: (x > 3).

3. Asymptote: (x = 3) Simple, but easy to overlook..

4. Intercepts:

   • x‑intercept: Set (f(x)=0) → (\log_2 (x-3) = -1) → (x-3 = 2^{-1}= \frac12) → (x = 3.5).
   • y‑intercept: Not applicable because (x=0) is outside the domain.

5. Reference points:

   • At (x = 4) (i.e., (h+1)): (f(4)=\log_2 1 +1 = 1).
   • At (x = 5) (i.e., (h+2)): (f(5)=\log_2 2 +1 = 2).
   • At (x = 7) (i.e., (h+4)): (f(7)=\log_2 4 +1 = 3).

6. Plot: Draw the vertical asymptote at (x=3), mark the points (4,1), (5,2), (7,3), and the x‑intercept (3.5,0). Connect them with a smooth curve that rises to the right and approaches the asymptote from the left.

Problem 2: Graph (g(x)=\log_{1/3} (2x+4) - 2)

1. Parameters:

   • Base (b = \frac13) (0 < b < 1, so the graph falls).
   • Horizontal shift solved from (2x+4>0) → (x>-2); rewrite as (\log_{1/3}[2(x+2)]) → (h = -2) after factoring the 2 into the argument (the factor 2 does not affect the shift).
   • Vertical shift (k = -2).

2. Domain: (x > -2).

3. Asymptote: (x = -2).

4. Intercepts:

   • x‑intercept: Set (g(x)=0) → (\log_{1/3} (2x+4) = 2) → (2x+4 = \left(\frac13\right)^2 = \frac19) → (2x = \frac19 - 4 = \frac19 - \frac{36}{9} = -\frac{35}{9}) → (x = -\frac{35}{18}\approx -1.944).
   • y‑intercept: Plug (x=0): (g(0)=\log_{1/3} 4 - 2). Compute (\log_{1/3}4 = \frac{\ln 4}{\ln (1/3)} \approx \frac{1.386}{-1.099}= -1.26). Thus (g(0)\approx -1.26-2 = -3.26). So the point (0, -3.26) lies on the curve.

5. Reference points (using base 1/3):

   • At (x = -1) (argument = 2·(-1)+4 = 2): (g(-1)=\log_{1/3}2 -2). Approximate (\log_{1/3}2 = \frac{\ln 2}{\ln (1/3)}\approx \frac{0.693}{-1.099}= -0.63). So (g(-1)\approx -2.63).
   • At (x = -0.5) (argument = 3): (g(-0.5)=\log_{1/3}3 -2\approx \frac{1.099}{-1.099}-2 = -1-2 = -3).

6. Plot: Draw the vertical asymptote at (x=-2). Because the base is less than 1, the curve approaches the asymptote from above on the left and falls to the right, passing through the calculated points and the x‑intercept near (-1.94) Not complicated — just consistent..

Problem 3: Determine the graph of (h(x)=\log_5 (x^2-4x+4))

1. Simplify the argument:
   (x^2-4x+4 = (x-2)^2).

2. Rewrite: (h(x)=\log_5 [(x-2)^2]). Since the argument is a square, it is always non‑negative; however, a logarithm requires a strictly positive argument, so we need ((x-2)^2>0) → (x\neq 2).

3. Domain: All real numbers except (x=2).

4. Asymptote: Because the argument never reaches zero, there is no vertical asymptote; instead, the graph has a hole at (x=2).

5. Behavior:

   • For (x>2), ((x-2)^2) grows, so (\log_5) of it increases slowly.
   • For (x<2), the square is still positive, and the log behaves symmetrically because ((x-2)^2) is the same for distances equidistant from 2.

6. Key points:

   • At (x=3): argument = 1 → (\log_5 1 = 0). So (3, 0).
   • At (x=1): argument = 1 → (1, 0).
   • At (x=4): argument = 4 → (\log_5 4 \approx 0.861).
   • At (x=0): argument = 4 → same value, (0, 0.861).

7. Plot: Draw a symmetric curve about the vertical line (x=2) that dips toward (-\infty) as (x) approaches 2 from either side (since (\log_5) of a number approaching 0 is (-\infty)). The graph passes through (1,0) and (3,0) and rises slowly for larger (|x-2|) Turns out it matters..

Problem 4: Find the inverse of (f(x)=\log_3 (2x-5) + 4) and graph it

1. Swap variables: Let (y = \log_3 (2x-5) + 4) Small thing, real impact..

2. Isolate the log: (y-4 = \log_3 (2x-5)).

3. Exponentiate: (3^{,y-4}=2x-5).

4. Solve for x: (x = \frac{3^{,y-4}+5}{2}).

5. Replace y with x for the inverse function:

[ f^{-1}(x)=\frac{3^{,x-4}+5}{2} ]

6. Graphing the inverse:
   • The inverse is an exponential function with base 3, shifted right by 4 units and scaled vertically by (\frac{1}{2}) then shifted up by (\frac{5}{2}).
   • Its domain is all real numbers; range is (x> \frac{5}{2}) (the original function’s domain).
   • Plot the original logarithmic graph (as in Problem 1) and reflect it across the line (y=x). The reflected points become:
     • Original x‑intercept (3.5, 0) → Inverse point (0, 3.5).
     • Original point (4, 1) → Inverse (1, 4).
   • Connect these reflected points with a smooth exponential curve.

Common Mistakes and How to Avoid Them

Frequently Asked Questions

Q1: How do I quickly find the vertical asymptote for a transformed logarithm?
A: Set the argument equal to zero and solve for (x). The solution is the asymptote (x = h). Any multiplicative constant inside the log does not affect the asymptote; only the additive shift does.

Q2: Can a logarithmic function have a horizontal asymptote?
A: No. Logarithmic functions extend infinitely in the vertical direction; they only possess a vertical asymptote Surprisingly effective..

Q3: Why does a base less than 1 make the graph fall to the right?
A: Because (\log_{b}(x)) with (0<b<1) is equivalent to (-\log_{1/b}(x)). The negative sign flips the growth direction, producing a decreasing curve.

Q4: What is the relationship between a logarithmic function and its inverse?
A: The inverse of a logarithmic function is an exponential function with the same base. Graphically, they are mirror images across the line (y = x) But it adds up..

Q5: How many points do I need to plot to get an accurate graph?
A: At minimum, plot the asymptote, intercept(s), and three reference points (e.g., (h+1), (h+b), (h+b^2)). More points improve accuracy, especially when the base is close to 0 or 1 Simple as that..

Tips for Efficient Homework Completion

1. Create a template on your notebook: a table with rows for domain, asymptote, intercepts, and reference points. Fill it in for each problem before drawing.
2. Use a calculator only for evaluating logs; keep the algebraic steps on paper to avoid accidental rounding errors.
3. Check symmetry: If the argument is a perfect square or involves ((x-h)^2), anticipate a symmetric graph about the line (x = h).
4. Label axes clearly with the asymptote line drawn as a dashed vertical line; this prevents misreading later.
5. Cross‑verify by reflecting the plotted points across (y = x) when you compute an inverse; the two graphs should match perfectly.

Conclusion

Graphing logarithmic functions in Unit 7, Homework 5 becomes manageable once you internalize the five core components: base behavior, domain/asymptote, intercepts, reference points, and transformation effects. By following the systematic steps illustrated above, you can produce accurate graphs, verify answers, and deepen your conceptual understanding—skills that will serve you well in higher‑level mathematics and real‑world applications. Keep this guide handy, practice with additional functions, and soon the process will feel as natural as plotting a straight line.