# How do you graph y=4x+3 using a table?

Jan 10, 2017

Well, think of it this way: you plug in something for $x$, and you follow the expression $4 x + 3$ by multiplying by $4$ then adding $3$, to get the corresponding output value $y$.

This equation is linear: $y = m x + b$,

where $m$ is the slope (change in y over change in x) and $b$ is the y-intercept (where it hits the $y$ axis).

Your equation just has $m = 4$ and $b = 3$.

So, let's make a quick table. You need at least two points to make a straight line, and three points to make a curve. Let's do five points though for practice.

$x = - 2 \to y = 4 \left(- 2\right) + 3 = - 5$
$x = - 1 \to y = 4 \left(- 1\right) + 3 = - 1$
$x = 0 \to y = 4 \left(0\right) + 3 = 3$
$x = 1 \to y = 4 \left(1\right) + 3 = 7$
$x = 2 \to y = 4 \left(2\right) + 3 = 11$

In a table it then looks like this:

$\textcolor{w h i t e}{\left[\begin{matrix}\textcolor{b l a c k}{\underline{y}} & \textcolor{b l a c k}{\underline{\text{|}}} & \textcolor{b l a c k}{\underline{x}} \\ \textcolor{b l a c k}{- 5} & \textcolor{b l a c k}{|} & \textcolor{b l a c k}{- 2} \\ \textcolor{b l a c k}{- 1} & \textcolor{b l a c k}{|} & \textcolor{b l a c k}{- 1} \\ \textcolor{b l a c k}{3} & \textcolor{b l a c k}{|} & \textcolor{b l a c k}{0} \\ \textcolor{b l a c k}{7} & \textcolor{b l a c k}{|} & \textcolor{b l a c k}{1} \\ \textcolor{b l a c k}{11} & \textcolor{b l a c k}{|} & \textcolor{b l a c k}{2}\end{matrix}\right]}$

And after plotting each point by looking at $x$ and $y$, and moving to the right $x$ units and up $y$ units (if negative, move the opposite direction), we get:

graph{4x + 3 [-2, 2, -5, 11]}

If you look on this graph, you should be able to find every point we had on the table.