Which graph shows the parent function f(x)=x2 horizontally compressed to form the graph of y=f(3x) ?(1 point) describe defining features

#1 has dots at (-1,9) and (1,9)
#2 has dots at (-3,1) and (1,3)
3# has dots at (-1, 4) and (1,4)
#4 has dots at (-1,1) and (1,1)

1 answer

To understand how the graph of \( f(x) = x^2 \) transforms into \( y = f(3x) \), we first recognize that compressing a graph horizontally involves changing the input to the function. Specifically, when you replace \( x \) with \( 3x \), it compresses the function by a factor of \( \frac{1}{3} \).

The parent function \( f(x) = x^2 \) produces points that can be calculated as follows:

  • \( f(-1) = (-1)^2 = 1 \)
  • \( f(0) = 0^2 = 0 \)
  • \( f(1) = 1^2 = 1 \)

The graph of \( f(x) = x^2 \) gives us the following points:

  • (-1, 1)
  • (0, 0)
  • (1, 1)

When we transform it to \( y = f(3x) = (3x)^2 = 9x^2 \), we need to identify how this affects specific \( x \) values.

To find the new \( y \) outputs at the same \( x \) values:

  • For \( x = -\frac{1}{3} \), \( f(3(-\frac{1}{3})) = f(-1) = 1 \) ⇒ point is (-1/3, 1)
  • For \( x = 0 \), \( f(3(0)) = f(0) = 0 \) ⇒ point is (0, 0)
  • For \( x = \frac{1}{3} \), \( f(3(\frac{1}{3})) = f(1) = 1 \) ⇒ point is (1/3, 1)

As we see, the \( y \)-values for \( x = -\frac{1}{3} \) and \( x = \frac{1}{3} \), both yield \( y = 1 \).

Now, let's look at the provided graphs to identify the one that has characteristics compatible with these transformations:

  1. Graph #1 has dots at (-1, 9) and (1, 9). These points are too high for the transformed function.
  2. Graph #2 has dots at (-3, 1) and (1, 3). These don't align with the computed points.
  3. Graph #3 has dots at (-1, 4) and (1, 4). Again, not compatible.
  4. Graph #4 has dots at (-1, 1) and (1, 1). This matches our transformed points when evaluating \( f(3x) \) for \( x = -\frac{1}{3} \) and \( x = \frac{1}{3} \).

The answer is Graph #4 which correctly reflects the horizontal compression resulting in a similar shape to the parent function \( f(x) = x^2 \), maintaining the correct height at \( y = 1 \) for these new \( x \) values.