Question

What is the equation in slope-intercept form of the line graphed below?

Answer 1

For this case we have that by definition, the equation of the line of the slope-intersection form is given by:

$y = mx + b$

Where:

m: It's the slope

b: It is the cut-off point with the y axis

According to the graph, we place two points through which the line passes:

$(x_ {1}, y_ {1}) :( 6,0)\\(x_ {2}, y_ {2}) :( 0, -3)$

We found the slope:

$m = \frac{y_ {2} -y_ {1}} {x_ {2} -x_ {1}}$

Substituting we have:

$m = \frac {-3-0} {0-6} = \frac {-3} {- 6} = \frac {1} {2}$

Thus, the equation is of the form:

$y = \frac {1} {2} x + b$

We substitute one of the points and find the cut-off point:

$0 = \frac {1} {2} (6) + b\\0 = 3 + b\\-3 = b\\b = -3$

Finally, the equation is:

$y = \frac {1} {2} x-3$

ANswer:

$y = \frac {1} {2} x-3$