Question

f(x) = x². what is g(x)? graph with f(x) (blue parabola) and g(x) (red parabola), point (2,1) on f(x). options: a. g(x) = (1/4)x², b. g(x) = 4x², c. g(x) = (1/2)x², d. g(x) = (1/4 x)²

Explanation:

Step1: Analyze the point on \( g(x) \)

We know that \( f(x)=x^2 \) and the graph of \( g(x) \) passes through the point \((2,1)\) (since it intersects or is related to the point shown, and we can test the options with \( x = 2 \)).

Step2: Test option A

For option A, \( g(x)=\frac{1}{4}x^2 \). Substitute \( x = 2 \): \( g(2)=\frac{1}{4}(2)^2=\frac{1}{4}\times4 = 1 \). This matches the \( y \)-value at \( x = 2 \) for \( g(x) \).

Step3: Test option B

For option B, \( g(x)=4x^2 \). Substitute \( x = 2 \): \( g(2)=4\times(2)^2=4\times4 = 16
eq1 \). So B is incorrect.

Step4: Test option C

For option C, \( g(x)=\frac{1}{2}x^2 \). Substitute \( x = 2 \): \( g(2)=\frac{1}{2}\times(2)^2=\frac{1}{2}\times4 = 2
eq1 \). So C is incorrect.

Step5: Test option D

For option D, \( g(x)=(\frac{1}{4}x)^2=\frac{1}{16}x^2 \). Substitute \( x = 2 \): \( g(2)=\frac{1}{16}\times(2)^2=\frac{1}{16}\times4=\frac{1}{4}
eq1 \). So D is incorrect.

Answer:

A. \( g(x)=\frac{1}{4}x^2 \)