To solve this problem, we'll follow these steps:
- Step 1: Analyze the graph provided. The graph appears to have a distinct 'V' shape, indicating that at least part of the graph is reflected about the x-axis.
- Step 2: Recognize that a standard quadratic function f(x)=x2−3 would have a vertex at (0, -3) if unaltered. Any sections of this below the x-axis being reflected upwards suggests the graph of an absolute value function.
- Step 3: Apply the absolute value function: When the function is f(x)=∣x2−3∣, all values of f(x) which are below the x-axis due to x2 terms less than 3 are reflected above the x-axis.
We can eliminate any function that does not involve an absolute value or shift consistent with reflected x-intercepts. Comparing with each choice:
- Choice 1 f(x)=x2−3 is incorrect because the graph wouldn't show the V-like reflection above the x-axis.
- Choice 2 f(x)=∣x2−3∣ matches perfectly since this transformation reflects all negative portions of the x2−3 graph above the x-axis as observed.
- Choice 3 f(x)=∣x2+3∣ is incorrect as this has no crossing or vertex shifts indicating a reflection starting point below the x-axis.
- Choice 4 f(x)=x2 doesn't reflect along y-values shifted down from 3.
Therefore, the function that corresponds to the graph is f(x)=∣x2−3∣.
f(x)=∣x2−3∣