Question

State the equation, in slope-intercept form, of each of the following graphs of linear relations. Explain how the equation was determined.

Answer 1

The equation, in slope-intercept form, of each of the following graphs of linear relations are:

1. y = 1/2(x) - 1.

2. y = -3x

3. y = 16x + 400

In Mathematics and Euclidean Geometry, the intercept form of the equation of a standard line can be modeled by the following formula;

x/a + y/b = 1

Where:

a and b are x-intercept and y-intercept respectively.

Part 1

Since the graph of the straight line function has an x-intercept at (2, 0) and a y-intercept at (0, -1), an equation that represents the line can be written as follows;

x/2 - y/1 = 1

(x - 2y)/2 = 1

x - 2y = 2

y = x/2 - 2/2

y = 1/2(x) - 1

Part 2

Since the graph pass through the origin (0, 0), it represents a proportional relationship;

y = mx

y = (3/-1)x

y = -3x

Part 3.

At data point (0, 400) and a slope of 16, an equation for this line can be calculated by using the point-slope form as follows:

`y - y_1 = m(x - x_1)`

y - 400 = 16(x - 0)

y = 16x + 400

Answer 2

slope intercept form is: y=mx+b
so to find the slope select two points that are on the line than by the slope formula (y2-y1/x2-x1) find the slope. then to find b select one point from the line and substitute x and y with the values of your point then solve the equation and you will find the b. to finish, write y= (the slope that you found)x plus or minus (the b that you found)