If we know the slope and the \(y\)-intercept of a linear relation, we can use this information to graph the relation.

Solution

• We know the slope,  \(m=4\).
  This means the rise is equal to \(4\) and the run is equal to \(1\).
• We know the \(y\)-intercept,  \(b=-1\)
• A positive rise is represented by moving up vertically.  A positive run is represented by moving right horizontally.

Solution

• First plot the \(y\)-intercept. We know in this example the point is \((0,-1)\).
• Use a ruler to connect the points and extend the line to represent all possible points that satisfy the linear relation.

Solution

We know the slope,  \(m=-\dfrac{2}{3}\).

Therefore,

• we can either graph this relation using a rise of \(-2\) and a run of \(3\), or
• we can use a rise of \(2\) and a run of \(-3\).

Both options will give us the same graph!

Solution

Approach 1

• First, let's use a rise of \(-2\) and a run of \(3\).
• We know the \(y\)-intercept, \(b=6\).
• Remember a positive rise is represented by moving up vertically. Because we have a negative rise in this example we will be moving down vertically.

Solution

Approach 1

• First, plot the \(y\)-intercept at point \((0,6)\).
• Use a ruler to connect the points and extend the line to represent all possible points that satisfy the linear relation.

Solution

Approach 2

• The rise is \(2\) and the run is \(-3\).
• A positive run is represented by moving right horizontally. Our negative run means we will be moving left horizontally.

Solution

• The equation is not in the form \(y=mx+b\).
• Use inverse operations and solve for \(y\) in terms of \(x\).

  \(3x-4y\)

  \(=-12\)

   

  \(-4y\)

  \(=-12-3x\)

   

  \(y\)

  \(=\dfrac{-12-3x}{-4}\)

   

  \(y\)

  \(=3+\dfrac{3}{4}x\)

   

  \(y\)

  \(=\dfrac{3}{4}x+3 \)

   

• We know the slope, \(m=\dfrac{3}{4}\). This means the rise is equal to \(3\) and the run is equal to \(4\).
• We know the \(y\)-intercept,  \(b=3\).

Solution

We now look at how to plot \(y=\dfrac{3}{4}x+3\).

• First plot the \(y\)-intercept at point \((0,3)\).
• Use a ruler to connect the points and extend the line to represent all possible points that satisfy the linear relation.