Question

scaling the cube root function quick check
what is the effect on the y-intercept in the graph of $f(x) = \sqrt3{x}$ when $f(x)$ is replaced by $f(5x)$? (1 point)
\bigcirc the y-intercept stretches horizontally by a factor of 5.
\bigcirc the y-intercept shifts up by a distance of 5.
\bigcirc the y-intercept compresses horizontally by a factor of 5.
\bigcirc the y-intercept does not change.

Explanation:

Step1: Find y-intercept of \( f(x) = \sqrt[3]{x} \)

To find the y - intercept, we set \( x = 0 \). Substituting \( x = 0 \) into \( f(x)=\sqrt[3]{x} \), we get \( f(0)=\sqrt[3]{0}=0 \). So the y - intercept of \( f(x)=\sqrt[3]{x} \) is \( (0,0) \).

Step2: Find y-intercept of \( f(5x)=\sqrt[3]{5x} \)

To find the y - intercept of \( y = f(5x)=\sqrt[3]{5x} \), we set \( x = 0 \). Substituting \( x = 0 \) into the function, we have \( y=\sqrt[3]{5\times0}=\sqrt[3]{0} = 0 \). So the y - intercept of \( f(5x) \) is also \( (0,0) \).

Step3: Analyze the change (or lack thereof) of y-intercept

We see that both \( f(x)=\sqrt[3]{x} \) and \( f(5x)=\sqrt[3]{5x} \) have a y - intercept at \( (0,0) \). A horizontal stretch or compression of a function \( y = f(x) \) to \( y=f(kx) \) (where \( k>0 \)) affects the x - values of the graph, but when we find the y - intercept, we set \( x = 0 \). Since substituting \( x = 0 \) into both functions gives the same y - value (0), the y - intercept does not change.

Answer:

The y - intercept does not change.