to find the equation of a linear equation given two points: 1) find the slope y=mx+b here's the equation: (y2-y1)/(x2-x1) y2- y value of the 2nd point y1- y value of 1st point x2- x value of 2nd point x1- x value of 1st point 1st point- (0, 12) 2nd point- (1,3) so your slope equation is this: (3-12) / (1-0) or -9/1 or -9 m is -9 2) find the y intercept y=mx+b b is whatever y is when x is 0. There's a long way to find it, but you already have it, because one of your points is (0,12). They told you what y was when x was 0. b is 12. your linear equation is y=-9x+12 to find the equation of an exponential equation given two points: 1) find a y=a(b)<span>× </span>if one of your points has a 0 as the x value, the y value is a. you do have a point like this: (0,12) so... <span>a is 12 </span>2) find b y=a(b)<span>× put what you know a is in the equation: y=12(b)</span><span>× then put the x and y values of the other point in for x and y in the equation so (1,3) 1 is x and 3 is y 3=12(b)^1 when the exponent is 1, it disappears: 3=12b simplify: b= 3/12 or b=1/4
then put all that in the equation: y=12(3/4)</span><span>×</span>
