SPM Practice 3 (Linear Law) – Question 2

Question 2 (10 marks):
Use a graph to answer this question.
Table shows the values of two variables, x and y, obtained from an experiment. A straight line will be obtained when a graph of $\frac{y^2}{x}\text{ against }\frac{1}{x}$ is plotted.


(a) Based on Table, construct a table for the values of $\frac{1}{x}\text{ and }\frac{y^2}{x}.$  
$\begin{array}{l}\left(b\right)\text{ Plot }\frac{y^2}{x}\text{ against }\frac{1}{x}\text{, using a scale of 2 cm to 0}\text{.1 unit on the }\frac{1}{x}\text{-axis}\\ \text{ and 2cm to 2 units on the }\frac{y^2}{x}\text{-axis}\text{.}\\ \text{ Hence, draw the line of best fit}\text{.}\end{array}$

(c) Using the graph in 11(b)
(i) find the value of y when x = 2.7,
(ii) express y in terms of x.

Solution:
(a)


(b)


(c)(i)
$\begin{array}{l}\text{When }x=2.7,\frac{1}{x}=0.37\\ \text{From graph,}\\ \frac{y^2}{x}=5.2\\ \frac{y^2}{2.7}=5.2\\ y=3.75\end{array}$


(c)(ii)
$\begin{array}{l}\text{Form graph, }y\text{-intercept, }c\text{ = –4}\\ \text{gradient, }m=\frac{16-\left(-4\right)}{0.8-0}=25\\ Y=mX+c\\ \frac{y^2}{x}=25\left(\frac{1}{x}\right)-4\\ y=\sqrt{25-4x}\end{array}$