Question

y = 2x^4 + 2
y = \frac{1}{2}x^3 + 2
y = \sqrt3{5x} + 2
y = \frac{1}{2}\sqrt{x} + 2

Explanation:

Identify key features of the graph

• The graph has a vertical intercept at \((0, 2)\).
• It exhibits a horizontal inflection point (flat spot) at \((0, 2)\).
• The graph extends from the third quadrant (bottom-left) to the first quadrant (top-right), which is characteristic of an odd-degree polynomial function with a positive leading coefficient, specifically a cubic function of the form \(y = ax^3 + k\).

Test key points on the options

• For \(y = \frac{1}{2}x^3 + 2\):
• At \(x = 0\), \(y = \frac{1}{2}(0)^3 + 2 = 2\). This matches the y-intercept \((0, 2)\).
• At \(x = 2\), \(y = \frac{1}{2}(2)^3 + 2 = 4 + 2 = 6\). Looking at the graph, at \(x = 2\), the curve passes exactly through \(y = 6\).
• At \(x = -2\), \(y = \frac{1}{2}(-2)^3 + 2 = -4 + 2 = -2\). Looking at the graph, at \(x = -2\), the curve passes exactly through \(y = -2\).

Eliminate incorrect options

• \(y = 2x^4 + 2\): This is an even-degree polynomial (quartic), which would have both ends pointing upwards (U-shaped), not matching the S-shape of the graph.
• \(y = \sqrt[3]{5x} + 2\): At \(x = 1\), \(y = \sqrt[3]{5} + 2 \approx 3.71\). At \(x = 1.6\), \(y = 2 + 2 = 4\). The graph shows \(y = 4\) occurs much further right, and the shape of a cube root function is flatter horizontally as \(x\) increases, whereas this graph grows rapidly.
• \(y = \frac{1}{2}\sqrt{x} + 2\): This is a square root function, which is only defined for \(x \ge 0\), but the graph clearly exists for negative values of \(x\).

Answer:

• (A) \(y = 2x^4 + 2\)
• (B) \(y = \frac{1}{2}x^3 + 2\) (Correct answer)
• (C) \(y = \sqrt[3]{5x} + 2\)
• (D) \(y = \frac{1}{2}\sqrt{x} + 2\)