Q:

Write the equation of the linear and exponential functions that pass through the points (0,12) and (1,3).Linear equation, y=mx+bExponential equation, y=a(b)×

Accepted Solution

A:
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)×
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...
a is 12
2) find b     y=a(b)×
put what you know a is in the equation: y=12(b)×
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)×