Question

Use the table to write a linear function that relates $\mathrm{y}$ to $\mathrm{x}$.

Answer 1

Step 1: Find the slope of the line using the points.

From the given table, Ordered pairs are $(-3,3),(0,5),(3,7),(6,9)$.

We know, slope of the line is $\mathrm{m}=\frac{\mathrm{change}\mathrm{in}\mathrm{y}}{\mathrm{change}\mathrm{in}\mathrm{x}}=\frac{{\mathrm{y}}_2-{\mathrm{y}}_1}{{\mathrm{x}}_2-{\mathrm{x}}_1}$

Now using the two points $(-3,3),(0,5)$, here ${\mathrm{x}}_1=-3,{\mathrm{x}}_2=0$ and ${\mathrm{y}}_1=3,{\mathrm{y}}_2=5$

$$\begin{gathered} \therefore \mathrm{m}=\frac{5-3}{0-(-3)} \\ =\frac{2}{3} \end{gathered}$$

Step 2: Finding the linear function that relates $\mathrm{y}$ to $\mathrm{x}$.

We know, Linear function is of the form $\mathrm{y}=\mathrm{mx}+\mathrm{c}$

$\Rightarrow \mathrm{y}=\frac{2}{3}\mathrm{x}+\mathrm{c}$ $\left[\because \mathrm{m}=\frac{2}{3}\right]$

$\Rightarrow 5=\frac{2}{3}\times 0+\mathrm{c}$ (Putting $\mathrm{y}=5$ and $\mathrm{x}=0$ i.e.,$(0,5)$)

$$\begin{gathered} \Rightarrow 5=\frac{2}{3}\times 0+\mathrm{c} \\ \Rightarrow 5=0+\mathrm{c} \\ \Rightarrow 5=\mathrm{c} \end{gathered}$$

Now again put $\mathrm{c}=5$ in linear function, we get $\mathrm{y}=\frac{2}{3}\mathrm{x}+5$.