MATH SOLVE

4 months ago

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)×

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)×