# How do you write the equation in function notation {(-3, -5) (-2, -3) (0, 1)}?

Jun 21, 2015

These three points are colinear and not vertical, so they can be written as a linear function: $f \left(x\right) = 2 x + 1$

If required, restrict the domain to $\left\{- 3 , - 2 , 0\right\}$ and the range to $\left\{- 5 , - 3 , 1\right\}$

#### Explanation:

If a line passes through two points $\left({x}_{1} , {y}_{1}\right)$ and $\left({x}_{2} , {y}_{2}\right)$ then the slope $m$ of the line is given by the formula:

$m = \frac{\Delta y}{\Delta x} = \frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}}$

In the case of points $\left(- 3 , - 5\right)$ and $\left(- 2 , - 3\right)$ we have

$m = \frac{- 3 - \left(- 5\right)}{- 2 - \left(- 3\right)} = \frac{2}{1} = 2$

In the case of points $\left(- 2 , - 3\right)$ and $\left(0 , 1\right)$ we have

$m = \frac{1 - \left(- 3\right)}{0 - \left(- 2\right)} = \frac{4}{2} = 2$

So these points lie along a line of slope $2$ with equation:

$f \left(x\right) = 2 x + c$ for some constant $c$

Since $\left(0 , 1\right)$ lies on the line $f \left(0\right) = 1$

But $f \left(0\right) = 2 \cdot 0 + c = 0 + c = c$

So $c = 1$ and the formula for $f \left(x\right)$ is $f \left(x\right) = 2 x + 1$