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Sagot :
To address the problem of determining the graph of the function [tex]\( g(x) = \frac{1}{4} f(x) \)[/tex] given that [tex]\( f(x) = x^2 \)[/tex], let's work through the steps methodically.
### Step-by-Step Solution
1. Understanding [tex]\( f(x) = x^2 \)[/tex]:
- [tex]\( f(x) = x^2 \)[/tex] is a parabolic function that opens upwards with its vertex at the origin (0,0).
- For various values of [tex]\( x \)[/tex], the function [tex]\( f(x) = x^2 \)[/tex] will have the following values:
- [tex]\( f(-2) = (-2)^2 = 4 \)[/tex]
- [tex]\( f(-1) = (-1)^2 = 1 \)[/tex]
- [tex]\( f(0) = 0^2 = 0 \)[/tex]
- [tex]\( f(1) = 1^2 = 1 \)[/tex]
- [tex]\( f(2) = 2^2 = 4 \)[/tex]
- So, the points on the graph of [tex]\( f(x) \)[/tex] are:
[tex]\[ (-2, 4), (-1, 1), (0, 0), (1, 1), (2, 4) \][/tex]
2. Now considering [tex]\( g(x) = \frac{1}{4} f(x) \)[/tex]:
- [tex]\( g(x) = \frac{1}{4} f(x) = \frac{1}{4} x^2 \)[/tex] means that each value of [tex]\( f(x) \)[/tex] is scaled by a factor of [tex]\( \frac{1}{4} \)[/tex].
- For the same values of [tex]\( x \)[/tex], we calculate [tex]\( g(x) \)[/tex]:
- [tex]\( g(-2) = \frac{1}{4} \cdot 4 = 1 \)[/tex]
- [tex]\( g(-1) = \frac{1}{4} \cdot 1 = 0.25 \)[/tex]
- [tex]\( g(0) = \frac{1}{4} \cdot 0 = 0 \)[/tex]
- [tex]\( g(1) = \frac{1}{4} \cdot 1 = 0.25 \)[/tex]
- [tex]\( g(2) = \frac{1}{4} \cdot 4 = 1 \)[/tex]
- So, the points on the graph of [tex]\( g(x) \)[/tex] are:
[tex]\[ (-2, 1), (-1, 0.25), (0, 0), (1, 0.25), (2, 1) \][/tex]
3. Interpreting the graph:
- The graph of [tex]\( g(x) \)[/tex] will also be a parabola opening upwards, but it will be wider compared to the graph of [tex]\( f(x) \)[/tex].
- The vertical stretch factor [tex]\( \frac{1}{4} \)[/tex] effectively compresses the parabola vertically.
4. Conclusion:
- The graph of [tex]\( g(x) = \frac{1}{4} x^2 \)[/tex] is a wider parabola than [tex]\( f(x) = x^2 \)[/tex], with key points at [tex]\( (-2, 1), (-1, 0.25), (0, 0), (1, 0.25), (2, 1) \)[/tex].
Therefore, the graph of [tex]\( g(x) \)[/tex] is a vertically compressed version of the graph of [tex]\( f(x) \)[/tex].
### Step-by-Step Solution
1. Understanding [tex]\( f(x) = x^2 \)[/tex]:
- [tex]\( f(x) = x^2 \)[/tex] is a parabolic function that opens upwards with its vertex at the origin (0,0).
- For various values of [tex]\( x \)[/tex], the function [tex]\( f(x) = x^2 \)[/tex] will have the following values:
- [tex]\( f(-2) = (-2)^2 = 4 \)[/tex]
- [tex]\( f(-1) = (-1)^2 = 1 \)[/tex]
- [tex]\( f(0) = 0^2 = 0 \)[/tex]
- [tex]\( f(1) = 1^2 = 1 \)[/tex]
- [tex]\( f(2) = 2^2 = 4 \)[/tex]
- So, the points on the graph of [tex]\( f(x) \)[/tex] are:
[tex]\[ (-2, 4), (-1, 1), (0, 0), (1, 1), (2, 4) \][/tex]
2. Now considering [tex]\( g(x) = \frac{1}{4} f(x) \)[/tex]:
- [tex]\( g(x) = \frac{1}{4} f(x) = \frac{1}{4} x^2 \)[/tex] means that each value of [tex]\( f(x) \)[/tex] is scaled by a factor of [tex]\( \frac{1}{4} \)[/tex].
- For the same values of [tex]\( x \)[/tex], we calculate [tex]\( g(x) \)[/tex]:
- [tex]\( g(-2) = \frac{1}{4} \cdot 4 = 1 \)[/tex]
- [tex]\( g(-1) = \frac{1}{4} \cdot 1 = 0.25 \)[/tex]
- [tex]\( g(0) = \frac{1}{4} \cdot 0 = 0 \)[/tex]
- [tex]\( g(1) = \frac{1}{4} \cdot 1 = 0.25 \)[/tex]
- [tex]\( g(2) = \frac{1}{4} \cdot 4 = 1 \)[/tex]
- So, the points on the graph of [tex]\( g(x) \)[/tex] are:
[tex]\[ (-2, 1), (-1, 0.25), (0, 0), (1, 0.25), (2, 1) \][/tex]
3. Interpreting the graph:
- The graph of [tex]\( g(x) \)[/tex] will also be a parabola opening upwards, but it will be wider compared to the graph of [tex]\( f(x) \)[/tex].
- The vertical stretch factor [tex]\( \frac{1}{4} \)[/tex] effectively compresses the parabola vertically.
4. Conclusion:
- The graph of [tex]\( g(x) = \frac{1}{4} x^2 \)[/tex] is a wider parabola than [tex]\( f(x) = x^2 \)[/tex], with key points at [tex]\( (-2, 1), (-1, 0.25), (0, 0), (1, 0.25), (2, 1) \)[/tex].
Therefore, the graph of [tex]\( g(x) \)[/tex] is a vertically compressed version of the graph of [tex]\( f(x) \)[/tex].
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