Big Ideas Math Algebra 1, 2015
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Big Ideas Math Algebra 1, 2015 View details
2. Writing Equations in Point-Slope Form
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Exercise 35 Page 186

One method is by using the point and slope that we can identify from the point-slope form.

See solution.

Practice makes perfect

Let's recall the point-slope form of a linear function. y- y_1= m(x- x_1) Here m is the slope and ( x_1, y_1) is a point on the line. We will now explore two methods to graph a line given in point-slope form.

First Method

First we need to identify the slope and the point in the given function.

y- y_1&= m(x- x_1) [0.8em] y- 1&= 3/2(x- 4) We identified the point ( 4, 1) and a slope of 32. Recall that the slope is the change in y divided by the change in x. m= 3/2 ⇓ Δ y/Δ x= 3/2 We will now first plot the point ( 4, 1). Then we will, as indicated by the slope, take a step of 2 units to the right and 3 units up to find a second point.

The point (4,1) along with a slope with three vertical units and two horizontal units used to find the next point on a line with slope 3/2

By drawing a line through the points we have our graph.

The point (4,1) with label and the next point (6,4) with no label and a line drawn that passes through both points.

Second Method

The second method is to use the given equation to find a second point. This is done by substituting an arbitrary y- or x-coordinate into the equation and solving for the other variable. Let's substitute x= 8 and solve for y.
y-1=3/2(x-4)
y-1=3/2( 8-4)
â–Ľ
Solve for y
y-1=3/2(4)
y-1=12/2
y-1=6
y=7
We have found that the point (8,7) is on the graph. Now we can plot our two points and connect them with a line to graph our function.
The points (4,1) and (8,7) with a line drawn through them.