Question

Determine the equation of the transformed image after the transformations described are applied to the given graph. a) The graph of \(y=2 \log _{5} x-7\) is reflected in the \(x\) -axis and translated 6 units up. b) The graph of \(y=\log (6(x-3))\) is stretched horizontally about the \(y\) -axis by a factor of 3 and translated 9 units left.

Short answer

a) The transformed equation is \( y = -2 \log_5(x) + 13 \). b) The transformed equation is \( y = \log(2x) \).

Step-by-step solution

Identify the Original Functions and Transformations

For part (a), the original function is \( y = 2 \log_5(x) - 7 \). For part (b), the original function is \( y = \log (6(x-3)) \). We need to apply the given transformations to these functions.

Reflect and Translate the First Function (a)

Reflecting the function in the \( x \)-axis changes the sign of the function: \( y' = -(2 \log_5(x) - 7) \). Simplifying, we get: \( y' = -2 \log_5(x) + 7 \). Then, translating this function 6 units up means adding 6 to the equation, which gives: \( y'' = -2 \log_5(x) + 7 + 6 \). The transformed equation for part (a) is \( y = -2 \log_5(x) + 13 \).

Stretch and Translate the Second Function (b)

The horizontal stretch by a factor of 3 means replacing \( x \) in the function with \( \frac{x}{3} \). So, \( y' = \log (6( \frac{x}{3} - 3)) \). This simplifies to \( y' = \log (6( \frac{x-9}{3})) \) or \( y' = \log (2(x-9)) \). Translating this function 9 units left means replacing \( x \) with \( x + 9 \), which gives: \( y'' = \log (2((x + 9) - 9)) = \log (2x) \). The transformed equation for part (b) is \( y = \log(2x) \).

Key concepts

reflections in the x-axis

Reflections are a type of transformation that mirror a graph over a specific axis. Reflecting a function in the x-axis changes the sign of the output values (y-values). For instance, reflecting the graph of the function \( y = f(x) \) in the x-axis transforms it into \( y = -f(x) \). This means that every point on the graph will have its y-coordinate multiplied by -1. For example, if you have a point (a, b) on the original graph, it will become (a, -b) after the reflection. In our original exercise, the function \( y = 2 \log_5(x) - 7 \) is reflected in the x-axis to become \( y = -2 \log_5(x) + 7 \). Notice how every value is inverted.

horizontal stretches

Horizontal stretches (or compressions) change the width of a graph. This means you either 'stretch' it out wider or 'compress' it tighter horizontally.

If you need to stretch the graph horizontally by a factor of k, you replace every x in the function with \( \frac{x}{k} \). If k > 1, the graph stretches wider; if 0 < k < 1, it compresses. For example, if you have a function like \(y = f(x) \), a horizontal stretch by a factor of 3 changes it to \( y = f \left( \frac{x}{3} \right) \).

Look at part (b) of our exercise: the original function is \( y = \log(6(x-3)) \). A horizontal stretch by a factor of 3 changes it to \( y = \log(6 \left( \frac{x}{3} - 3 \right)) \), which simplifies to \( y = \log \left( 2(x-9) \right) \).

function translation

Translations are shifts of the graph either horizontally, vertically, or both. They do not alter the shape of the graph, merely its position. When translating a function:

• To move it up or down, add or subtract a number to/from the function. For instance, moving \( y = f(x) \) up by c units makes it \( y = f(x) + c \).
• To shift it left or right, add or subtract inside the function's argument. Moving \( y = f(x) \) left by c units changes the function to \( y = f(x + c) \), while moving it right by c units changes it to \( y = f(x - c) \).

For part (a) of our exercise: after reflecting the function \( y = 2 \log_5(x) - 7 \) in the x-axis, we get \( y = -2 \log_5(x) + 7 \). Translating it 6 units up gives us \( y = -2 \log_5(x) + 13 \).

In part (b): translating \( y = \log(2(x-9)) \) 9 units left yields \( y = \log(2(x+9 - 9)) = \log(2x) \), because adding 9 inside shifts the entire function left.