Answer :
To find the graph of the function [tex]\( g(x) = f(4x) \)[/tex] where [tex]\( f(x) = x^2 \)[/tex], let's go through the process step by step:
1. Understand the function [tex]\( f(x) \)[/tex]:
- The function [tex]\( f(x) = x^2 \)[/tex] is a standard quadratic function. Its graph is a parabola that opens upwards with its vertex at the origin (0,0).
2. Substitute [tex]\( 4x \)[/tex] into [tex]\( f(x) \)[/tex]:
- [tex]\( g(x) = f(4x) \)[/tex].
- Since [tex]\( f(x) = x^2 \)[/tex], we replace [tex]\( x \)[/tex] with [tex]\( 4x \)[/tex] in the function [tex]\( f(x) \)[/tex]. So, [tex]\( f(4x) = (4x)^2 \)[/tex].
3. Simplify [tex]\( g(x) \)[/tex]:
- When we simplify [tex]\( (4x)^2 \)[/tex]:
[tex]\[ g(x) = (4x)^2 = 16x^2 \][/tex]
Therefore, the function [tex]\( g(x) = 16x^2 \)[/tex].
4. Analyze the Transformation:
- The original function [tex]\( f(x) = x^2 \)[/tex] has been transformed into [tex]\( g(x) = 16x^2 \)[/tex].
- This transformation involves a horizontal compression by a factor of 4. To see why, note that [tex]\( x \)[/tex] in the original function [tex]\( f \)[/tex] is replaced by [tex]\( 4x \)[/tex]. In terms of the graph, every point [tex]\( (x, y) \)[/tex] on the graph of [tex]\( f \)[/tex] will be mapped to [tex]\( (\frac{x}{4}, y) \)[/tex] on the graph of [tex]\( g \)[/tex].
- Another way to see this is that [tex]\( f(4x) \)[/tex] will produce the same output as [tex]\( f(x) \)[/tex] but for an input that is [tex]\( \frac{1}{4} \)[/tex] of the original [tex]\( x \)[/tex].
5. Sketch the Graph:
- The graph of [tex]\( g(x) = 16x^2 \)[/tex] will still be a parabola opening upwards.
- However, compared to [tex]\( f(x) \)[/tex], it will be much narrower (compressed horizontally) due to the factor of 16.
So, summarizing:
- The original graph [tex]\( f(x) = x^2 \)[/tex] is transformed into [tex]\( g(x) = 16x^2 \)[/tex].
- The new graph will be a vertically stretched and horizontally compressed parabola compared to [tex]\( f(x) = x^2 \)[/tex].
Therefore, the graph of [tex]\( g(x) \)[/tex] is a parabola which is narrower than the graph of [tex]\( f(x) = x^2 \)[/tex] and has the equation [tex]\( y = 16x^2 \)[/tex].
1. Understand the function [tex]\( f(x) \)[/tex]:
- The function [tex]\( f(x) = x^2 \)[/tex] is a standard quadratic function. Its graph is a parabola that opens upwards with its vertex at the origin (0,0).
2. Substitute [tex]\( 4x \)[/tex] into [tex]\( f(x) \)[/tex]:
- [tex]\( g(x) = f(4x) \)[/tex].
- Since [tex]\( f(x) = x^2 \)[/tex], we replace [tex]\( x \)[/tex] with [tex]\( 4x \)[/tex] in the function [tex]\( f(x) \)[/tex]. So, [tex]\( f(4x) = (4x)^2 \)[/tex].
3. Simplify [tex]\( g(x) \)[/tex]:
- When we simplify [tex]\( (4x)^2 \)[/tex]:
[tex]\[ g(x) = (4x)^2 = 16x^2 \][/tex]
Therefore, the function [tex]\( g(x) = 16x^2 \)[/tex].
4. Analyze the Transformation:
- The original function [tex]\( f(x) = x^2 \)[/tex] has been transformed into [tex]\( g(x) = 16x^2 \)[/tex].
- This transformation involves a horizontal compression by a factor of 4. To see why, note that [tex]\( x \)[/tex] in the original function [tex]\( f \)[/tex] is replaced by [tex]\( 4x \)[/tex]. In terms of the graph, every point [tex]\( (x, y) \)[/tex] on the graph of [tex]\( f \)[/tex] will be mapped to [tex]\( (\frac{x}{4}, y) \)[/tex] on the graph of [tex]\( g \)[/tex].
- Another way to see this is that [tex]\( f(4x) \)[/tex] will produce the same output as [tex]\( f(x) \)[/tex] but for an input that is [tex]\( \frac{1}{4} \)[/tex] of the original [tex]\( x \)[/tex].
5. Sketch the Graph:
- The graph of [tex]\( g(x) = 16x^2 \)[/tex] will still be a parabola opening upwards.
- However, compared to [tex]\( f(x) \)[/tex], it will be much narrower (compressed horizontally) due to the factor of 16.
So, summarizing:
- The original graph [tex]\( f(x) = x^2 \)[/tex] is transformed into [tex]\( g(x) = 16x^2 \)[/tex].
- The new graph will be a vertically stretched and horizontally compressed parabola compared to [tex]\( f(x) = x^2 \)[/tex].
Therefore, the graph of [tex]\( g(x) \)[/tex] is a parabola which is narrower than the graph of [tex]\( f(x) = x^2 \)[/tex] and has the equation [tex]\( y = 16x^2 \)[/tex].
