Question

which is true about the graph of $k(x) = \frac{1}{2}\sqrt3{-x} + 1$?
i has one x -intercept and no y -intercept.
i has one x -intercept and one y -intercept.
i has two x -intercepts and one y -intercept.
i has no x -intercept and one y -intercept.

Explanation:

Step1: Find x-intercept (set k(x)=0)

Set $\frac{1}{2}\sqrt[3]{-x} + 1 = 0$

Step2: Isolate the radical term

$\frac{1}{2}\sqrt[3]{-x} = -1$

Step3: Eliminate the coefficient

$\sqrt[3]{-x} = -2$

Step4: Cube both sides

$-x = (-2)^3 = -8$

Step5: Solve for x

$x = 8$

Step6: Find y-intercept (set x=0)

$k(0) = \frac{1}{2}\sqrt[3]{-0} + 1 = 0 + 1 = 1$

Answer:

B. I has one x-intercept and one y-intercept.