Question

algebra 2a semester online practice
complete this assessment to review what youve learned. it will not count toward your grade.

a coordinate on the graph of $f(x) = \sqrt3{x}$ is $(27, 3)$. the function is transformed into $f(x) = k\sqrt3{x}$, and a point on the new graph is $(27, 12)$. what is the value of $k$? (1 point)

\bigcirc 9
\bigcirc 4
\bigcirc 12
\bigcirc $\frac{1}{4}$

Explanation:

Step1: Substitute the point into the new function

We know the new function is \( f(x) = k\sqrt[3]{x} \), and the point \((27, 12)\) lies on its graph. So we substitute \( x = 27 \) and \( f(x)=12 \) into the function: \( 12 = k\sqrt[3]{27} \).

Step2: Simplify the cube root

Calculate \( \sqrt[3]{27} \). Since \( 3^3 = 27 \), we have \( \sqrt[3]{27}=3 \). So the equation becomes \( 12 = k\times3 \).

Step3: Solve for \( k \)

To find \( k \), divide both sides of the equation \( 12 = 3k \) by 3: \( k=\frac{12}{3}=4 \).

Answer:

4 (corresponding to the option "4")