Problem
The phrase “how do i graph y” often appears when learning to graph an equation where \(y\) depends on \(x\). Consider the linear equation \(y = 2\cdot x - 1\). The task is to draw its graph on the coordinate plane.
Step 1: Recognize slope–intercept form
A line written as \(y = m\cdot x + b\) has:
- Slope \(m\): the rise per unit run (change in \(y\) per change in \(x\)).
- Y-intercept \(b\): the value of \(y\) when \(x=0\), i.e., the point \((0,b)\).
For \(y = 2\cdot x - 1\), the slope is \(m=2\) and the y-intercept is \(b=-1\).
Step 2: Plot the y-intercept
Set \(x=0\): \[ y = 2\cdot 0 - 1 = -1. \] So the graph passes through \((0,-1)\).
Step 3: Use the slope to find a second point
The slope \(m=2\) means: \[ \frac{\Delta y}{\Delta x} = 2 = \frac{2}{1}. \] So from any point on the line, moving \(1\) unit to the right (run \(+1\)) moves \(2\) units up (rise \(+2\)).
Starting at \((0,-1)\), move right \(1\) and up \(2\) to reach \((1,1)\).
Step 4: Draw the line through the points
Plot \((0,-1)\) and \((1,1)\), then draw the straight line passing through both points and extend it in both directions.
Verification using a small table of values
Another reliable way to graph \(y\) is to compute several \((x,y)\) pairs and plot them.
| \(x\) | \(y = 2\cdot x - 1\) |
|---|---|
| \(-1\) | \(-3\) |
| \(0\) | \(-1\) |
| \(1\) | \(1\) |
| \(2\) | \(3\) |
Result (graphing recipe): plot \((0,-1)\), use slope \(2\) to get a second point such as \((1,1)\), and draw the unique line through the points.
Visualization: graph of \(y = 2\cdot x - 1\)
Quick correctness check
Substituting a plotted point into the equation confirms the graph: for \((1,1)\), \[ 1 = 2\cdot 1 - 1, \] which is true, so the point lies on the line.