What is the effect on the $$ y $$-intercept in the graph of $$ f(x)=\sqrt[3]{x} $$ when $$ f(x) $$ is replaced by $$ f(x)+5 $$ ? (1 point) The $$ y $$-intercept shifts to the right by a distance of 5 . The $$ y $$-intercept shifts down by a distance of 5 . The $$ y $$-intercept shifts to the left by a distance of 5 . The $$ y $$-intercept shifts up by a distance of 5 .

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When you replace $$ f(x) = \sqrt[3]{x} $$ with $$ f(x) + 5 $$, the new function becomes $$ \sqrt[3]{x} + 5 $$.

- **Original y-intercept:** For $$ f(x) = \sqrt[3]{x} $$, the y-intercept is at $$ (0, 0) $$.
- **New y-intercept:** For $$ f(x) + 5 = \sqrt[3]{x} + 5 $$, when $$ x = 0 $$, the y-intercept is $$ 0 + 5 = 5 $$.

This means the y-intercept moves from $$ (0, 0) $$ to $$ (0, 5) $$, which is a shift **upwards** by 5 units.

**Answer:** The $$ y $$-intercept shifts up by a distance of 5.
The $$ y $$-intercept shifts up by a distance of 5.

What is the effect on the -intercept in the graph of when is
replaced by ? (1 point)
The -intercept shifts to the right by a distance of 5 .
The -intercept shifts down by a distance of 5 .
The -intercept shifts to the left by a distance of 5 .
The -intercept shifts up by a distance of 5 .

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What is the effect on the -intercept in the graph of when is replaced by ? (1 point) The -intercept shifts to the right by a distance of 5 . The -intercept shifts down by a distance of 5 . The -intercept shifts to the left by a distance of 5 . The -intercept shifts up by a distance of 5 .