Question

analyzing graphs of functions which graph shows a function where f(2) = 4? four graphs with radio buttons, one marked with x

Explanation:

To determine which graph shows \( f(2) = 4 \), we analyze the coordinates:

Step 1: Understand \( f(2) = 4 \)

This means when \( x = 2 \), the function's \( y \)-value (output) is \( 4 \). So we need a graph where the point \( (2, 4) \) lies on the function’s curve.

Step 2: Analyze each graph (conceptually, since we infer the first graph’s shape):

• The first graph (top) is a parabola opening upwards with vertex at the origin \((0,0)\), but shifted? Wait, no—wait, the first graph’s curve: let's check \( x = 2 \). If the parabola is \( y = x^2 \)? No, wait, the first graph’s \( y \)-axis has positive values. Wait, actually, the first graph (top) has a vertex at \((0,0)\) but the curve: when \( x = 2 \), what’s \( y \)? Wait, no—wait, the first graph (top) is a parabola opening upwards, but let's re-express. Wait, the first graph (top) has the curve starting at the origin and opening upwards? No, wait, the first graph’s \( y \)-axis: the orange curve is on the positive \( y \)-side. Wait, actually, the key is: the first graph (top) when \( x = 2 \), the \( y \)-value should be \( 4 \) if it’s \( y = x^2 \)? Wait, no—wait, \( f(2) = 4 \) means at \( x = 2 \), \( y = 4 \). Let’s check the first graph: if the grid is 1 unit per square, then at \( x = 2 \), moving up 4 units (since \( y = 4 \)) should lie on the curve. The first graph (top) is a parabola that passes through \( (2, 4) \) (since \( 2^2 = 4 \), if it’s \( y = x^2 \), but wait, the vertex is at \( (0,0) \), so \( y = x^2 \) would have \( (2, 4) \) on it. Wait, but the first graph’s curve: let’s confirm. The other graphs: the second graph opens downward (negative \( y \)-values for positive \( x \)), so \( f(2) \) would be negative, not 4. The third and fourth graphs are shifted, but the first graph (top) is the one where \( (2, 4) \) lies on the curve.

Answer:

The top - most graph (the first one with the orange curve opening upwards, vertex at the origin, and passing through \((2, 4)\)) shows \( f(2)=4 \). (Assuming the first graph is the correct one based on the \( y = x^2 \) - like shape where \( x = 2 \) gives \( y = 4 \).)