Question

#22
a student graphed the linear parent function $f$ and $h(x) = \frac{1}{3}f(x)$.
which statement is true about graphs $f$ and $h$?
\bigcirc the graph of $h$ is steeper than the graph of $f$.
\bigcirc the graph of $h$ is less steep than the graph of $f$.
\bigcirc the $y$-intercept of the graph of $h$ is 3 units above the $y$-intercept of the graph of $f$.
\bigcirc the $y$-intercept of the graph of $h$ is 3 units below the $y$-intercept of the graph of $f$.
submit

Explanation:

Step1: Define linear parent function

The linear parent function is $f(x) = x$, so its slope $m_f = 1$, y-intercept $(0,0)$.

Step2: Find $h(x)$ and its slope

Calculate $h(x)=\frac{1}{3}f(x)=\frac{1}{3}x$. Its slope $m_h = \frac{1}{3}$.

Step3: Compare slopes

Since $\frac{1}{3} < 1$, $h(x)$ is less steep.

Step4: Compare y-intercepts

Both $f(x)$ and $h(x)$ have y-intercept $(0,0)$, so no vertical shift.

Answer:

The graph of $h$ is less steep than the graph of $f$.