Chapter 1.5, Problem 51E

Solution

To Graph:

Try changing the value of $k$ in the function $y={\left(\frac{3^k}{30}\left(x-65\right)\right)}^{\frac{1}{k}}+1$ between $1$ and $2$ .

To State:

What kind of scaling curve is obtained for $k=1$ .

To give opinion:

Whether the decision that $k=1.7$ gives the fairest GPAs.

The graph of $y={\left(\frac{3^k}{30}\left(x-65\right)\right)}^{\frac{1}{k}}+1$ for $k=1$ is seen to be a straight line passing through $\left(65,1\right)$ and $\left(95,4\right)$ . The equation reduces to the linear equation $y=\frac{x}{10}-5.5$ .

For $k=1.7$ the curve is fairest because it gives higher than linear equivalent grades to percentages between $65$ and $95$ but trims grades lower than linear for percentages above $95$ . The graph looks pretty balanced around the linear equation $y=\frac{x}{10}-5.5$ .

Given:

The function $y={\left(\frac{3^k}{30}\left(x-65\right)\right)}^{\frac{1}{k}}+1$ where $k$ lies between $1$ and $2$ .

Concepts Used:

Graphing functions using a Graphing Calculator.

Graphs:

  

  

  

  

The graphs of the function $y={\left(\frac{3^k}{30}\left(x-65\right)\right)}^{\frac{1}{k}}+1$ where various values of $k$ are tried starting from $1$ going upto $2$ with interval of $0.1$ between the values.

Conclusion:

The graph of $y={\left(\frac{3^k}{30}\left(x-65\right)\right)}^{\frac{1}{k}}+1$ for $k=1$ is seen to be a straight line passing through $\left(65,1\right)$ and $\left(95,4\right)$ . The equation reduces to the linear equation $y=\frac{x}{10}-5.5$ .

For $k=1.7$ the curve is fairest because it gives higher than linear equivalent grades to percentages between $65$ and $95$ but trims grades lower than linear for percentages above $95$ . The graph looks pretty balanced around the linear equation $y=\frac{x}{10}-5.5$ .