Math Problem Statement

solve

If f(x) = 2x^2 - 4, give the function g which shows the graph of f(x) after: (a) translation by (1,0) followed by a horizontal stretch of scale factor 1/4.

Solution

We are given the function f(x)=2x24f(x) = 2x^2 - 4 and are asked to find a new function g(x)g(x) that represents the graph of f(x)f(x) after the following transformations:

  1. Translation by (10)\begin{pmatrix} 1 \\ 0 \end{pmatrix} (right by 1 unit).
  2. Horizontal stretch by a factor of 14\frac{1}{4}.

Step 1: Translation by (10)\begin{pmatrix} 1 \\ 0 \end{pmatrix}

A translation to the right by 1 unit is applied to the xx-coordinates. This changes the function from f(x)f(x) to f(x1)f(x-1).

So, the translated function is: f(x1)=2(x1)24f(x-1) = 2(x-1)^2 - 4 Now, let's expand (x1)2(x-1)^2: (x1)2=x22x+1(x-1)^2 = x^2 - 2x + 1 Thus, the translated function becomes: f(x1)=2(x22x+1)4=2x24x+24=2x24x2f(x-1) = 2(x^2 - 2x + 1) - 4 = 2x^2 - 4x + 2 - 4 = 2x^2 - 4x - 2

Step 2: Horizontal Stretch by a Factor of 14\frac{1}{4}

To apply a horizontal stretch by a factor of 14\frac{1}{4}, we replace xx in the function with x14\frac{x}{\frac{1}{4}} (i.e., multiply xx by 4).

The function after this transformation is: f(x41)=2(x41)24f\left( \frac{x}{4} - 1 \right) = 2\left( \frac{x}{4} - 1 \right)^2 - 4 Now expand (x41)2\left( \frac{x}{4} - 1 \right)^2: (x41)2=(x216x2+1)\left( \frac{x}{4} - 1 \right)^2 = \left( \frac{x^2}{16} - \frac{x}{2} + 1 \right) Substitute this back into the function: f(x41)=2(x216x2+1)4f\left( \frac{x}{4} - 1 \right) = 2 \left( \frac{x^2}{16} - \frac{x}{2} + 1 \right) - 4 Simplify the expression: =x28x+24=x28x2= \frac{x^2}{8} - x + 2 - 4 = \frac{x^2}{8} - x - 2

Thus, the function g(x)g(x), after both transformations, is: g(x)=x28x2g(x) = \frac{x^2}{8} - x - 2

Would you like more details on any of the steps?

Here are some related questions for further exploration:

  1. How would the graph change if we performed a vertical stretch instead of a horizontal one?
  2. What happens to the vertex of the parabola after the transformations?
  3. Can we apply a reflection instead of a stretch and find the new function?
  4. How does the horizontal stretch factor of 14\frac{1}{4} affect the graph's shape?
  5. What is the general process for applying multiple transformations to a function?

Tip: When applying transformations, always handle translations and stretches separately and in the correct order to avoid confusion.

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Math Problem Analysis

Mathematical Concepts

Transformations of Functions
Translations
Stretches and Compressions
Quadratic Functions

Formulas

Translation: f(x - h) for horizontal shift
Horizontal Stretch: f(kx) for stretch by factor 1/k

Theorems

Transformation Theorem

Suitable Grade Level

Grades 10-12