Question

this is a graph of which function?
a. (f(x) = x^2 + 4)
b. (f(x) = x^2 - 7)
c. (f(x) = \frac{1}{2}x^2 + 2)
d. (f(x) = -x^2 - 3)
e. (f(x) = -2x^2 + 4)

Explanation:

Identify parabola direction

The graph shows a parabola opening upward.
Using the Parabola Direction and Leading Coefficient knowledge points, an upward-opening parabola must have a positive leading coefficient \(a > 0\).
This rules out options d and e, which have negative leading coefficients.

Find the vertex

The vertex of the parabola is located on the y-axis below the x-axis.
Looking closely at the grid, the vertex is at \((0, -7)\).
Using the Quadratic Functions knowledge point, a quadratic function of the form \(f(x) = ax^2 + c\) has its vertex at \((0, c)\).
Since the vertex is below the x-axis, the y-intercept \(c\) must be negative.

Match with options

Let's evaluate the remaining options with positive leading coefficients:

• Option a: \(f(x) = x^2 + 4\) has a vertex at \((0, 4)\).
• Option b: \(f(x) = x^2 - 7\) has a vertex at \((0, -7)\).
• Option c: \(f(x) = \frac{1}{2}x^2 + 2\) has a vertex at \((0, 2)\).

The only function that matches a vertex below the x-axis at \((0, -7)\) is option b.

Answer:

• a. \(f(x) = x^2 + 4\)
• b. \(f(x) = x^2 - 7\) (Correct answer)
• c. \(f(x) = \frac{1}{2}x^2 + 2\)
• d. \(f(x) = -x^2 - 3\)
• e. \(f(x) = -2x^2 + 4\)