# How do you graph y=6x-3 by plotting points?

May 8, 2018

Plug in random values for $x$, then solve.

#### Explanation:

For example, let's pick $0$, $1$, and $2$.

$y = 6 \left(0\right) - 3 = - 3$

$y = 6 \left(1\right) - 3 = 3$

$y = 6 \left(2\right) - 3 = 9$

Then just plot $\left(0 , - 3\right)$, (1, 3), $\left(2 , 9\right)$ into a graph like so:

graph{y=6x-3 [-15, 15, -5, 10]}

May 8, 2018

See below

#### Explanation:

First of all, note that this function is a polynomial, because it can be written as a sum of powers of a variable with some coefficients:

$p \left(x\right) = {a}_{0} + {a}_{1} x + {a}_{2} {x}^{2} + \ldots + {a}_{n} {x}^{n}$

In your case, $n = 1$, ${a}_{0} = - 3$ and ${a}_{1} = 6$

And it is because $n = 1$ that this function is a line: every polyinomial of degree $1$ represents a line.

To draw a line, you only need two points, so that you can connect them. To sample two points from the equation, you need to choose two points ${x}_{1} , {x}_{2}$ and compute the respective images ${y}_{1} , {y}_{2}$. Then, draw them and connect the points $\left({x}_{1} , {y}_{1}\right)$ and $\left({x}_{2} , {y}_{2}\right)$

Of course, it makes no sense to choose "difficult" points to compute, since any couple will work. For this reason, you may choose, for example, ${x}_{1} = 0$ and ${x}_{2} = 1$, so that the computations will be easy:

${x}_{1} = 0 \setminus \implies {y}_{1} = 6 \setminus \cdot 0 - 3 = - 3 \setminus \implies {P}_{1} = \left(0 , - 3\right)$

${x}_{2} = 1 \setminus \implies {y}_{1} = 6 \setminus \cdot 1 - 3 = 3 \setminus \implies {P}_{2} = \left(1 , 3\right)$

Now you only need to draw ${P}_{1}$ and ${P}_{2}$ and connect them.