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Analyze the effect on the \( y \)-intercept in the graph of \( f(x)=\sqrt[3]{x} \) when \( f(x) \) is reptaced by \( f(x)+d \) and \( d \) is positive. (1 point) The \( y \)-intercept shifts up by a distance of \( d \). The \( y \)-intercept shifts to the right by a distance of \( d \). The \( y \)-intercept shifts to the left by a distance of \( d \). The \( y \)-intercept shifts down by a distance of \( d \).

Ask by Chambers Lawson. in the United States
Jan 23,2025

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Answer

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

Solution

When you replace \( f(x) = \sqrt[3]{x} \) with \( f(x) + d \), you're effectively adding a constant \( d \) to the entire function. The original \( y \)-intercept of \( f(x) = \sqrt[3]{x} \) is at \( (0, 0) \). After the transformation: \[ f(0) + d = \sqrt[3]{0} + d = 0 + d = d \] So, the new \( y \)-intercept is at \( (0, d) \), which means it has shifted **up** by a distance of \( d \). **Answer:** The \( y \)-intercept shifts up by a distance of \( d \).

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The Deep Dive

When you replace \( f(x) \) with \( f(x) + d \) where \( d \) is positive, the entire graph of the function shifts vertically upwards. This means that the original \( y \)-intercept, which is at \( (0,0) \) for \( f(x) = \sqrt[3]{x} \), will now be at \( (0, d) \). In simpler words, it’s like moving your function up the y-axis! So, the correct option is: The \( y \)-intercept shifts up by a distance of \( d \). Additionally, if you ever find yourself confused about how vertical and horizontal shifts affect a graph, just remember this: Adding a constant to the function (like \( +d \)) affects the \( y \)-values, while adding a constant to \( x \) inside the function shifts it horizontally. This simple rule can help you visualize any transformations quickly!

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