Question

shifting the cube root function quick check
analyze the effect on the y-intercept in the graph of $f(x) = \sqrt3{x}$ when $f(x)$ is replaced by $f(x) + d$ and $d$ is positive. (1 point)
\bigcirc the y-intercept shifts down by a distance of $d$
\bigcirc the y-intercept shifts to the right by a distance of $d$
\bigcirc the y-intercept shifts up by a distance of $d$
\bigcirc the y-intercept shifts to the left by a distance of $d$

Explanation:

To determine the effect on the \( y \)-intercept when \( f(x)=\sqrt[3]{x} \) is replaced by \( f(x)+d \) (where \( d>0 \)):

1. First, find the \( y \)-intercept of the original function \( f(x)=\sqrt[3]{x} \). The \( y \)-intercept occurs at \( x = 0 \), so \( f(0)=\sqrt[3]{0}=0 \). Thus, the original \( y \)-intercept is \( (0, 0) \).
2. For the new function \( f(x)+d=\sqrt[3]{x}+d \), find its \( y \)-intercept by setting \( x = 0 \): \( f(0)+d=\sqrt[3]{0}+d = 0 + d = d \). So the new \( y \)-intercept is \( (0, d) \).
3. Compare the original \( y \)-intercept \( (0, 0) \) and the new \( y \)-intercept \( (0, d) \). Since \( d>0 \), the \( y \)-coordinate increases by \( d \), which means the \( y \)-intercept shifts up by a distance of \( d \).

Answer:

C. The \( y \)-intercept shifts up by a distance of \( d \)