Question

scaling 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 $f(bx)$ and $b$ is greater than 1. (1 point)

• the y-intercept compresses horizontally by a factor of $b$
• the y-intercept stretches horizontally by a factor of $b$
• the y-intercept does not change
• the y-intercept shifts up by a distance of $b$

Explanation:

To determine the effect on the y - intercept of \(f(x)=\sqrt[3]{x}\) when \(f(x)\) is replaced by \(f(bx)\) (\(b > 1\)):

1. Recall the definition of the y - intercept: The y - intercept occurs when \(x = 0\).
2. For the original function \(f(x)=\sqrt[3]{x}\), the y - intercept is found by setting \(x = 0\), so \(f(0)=\sqrt[3]{0}=0\).
3. For the function \(f(bx)=\sqrt[3]{bx}\), find the y - intercept by setting \(x = 0\). Then \(f(b\times0)=\sqrt[3]{b\times0}=\sqrt[3]{0} = 0\).
4. Now, consider the horizontal transformation. The transformation from \(f(x)\) to \(f(bx)\) (\(b>1\)) is a horizontal compression by a factor of \(\frac{1}{b}\), but the y - intercept (the point \((0,0)\)) does not change its position because when \(x = 0\), \(bx=0\) regardless of the value of \(b\) (as long as \(b\) is a real number). The other options are incorrect:

• Option 1: The y - intercept is a single point \((0,0)\), and the concept of horizontal compression for a single point (the origin) does not apply in the way described. A horizontal compression of the graph affects the non - origin points, but the origin remains the same.
• Option 2: A horizontal stretch (which would occur when \(0 < b<1\)) is not what happens here, and again, the origin does not stretch.
• Option 4: There is no vertical shift (shifting up) because the transformation \(f(x)\to f(bx)\) is a horizontal transformation, not a vertical one.

Answer:

The y - intercept does not change.