Exponential Function

Question

Find the exponential equation whose graph passes through the points (-1,3/5 )and (2,75 )

y=

Answer 1

While Paul provides an exponential function, the exponential fuction desired, I believe, is y = ab^x

Then, as the function goes through (-1,3/5 ) and (2,75 ). we have

from the point going through (-1, 3/5) 3/5 = a b^-1 and

from the point going through (2, 75) 75 = a b²

Then, we divide the second expression by the first.

75/(3/5) = b²/b^-1

Remember that you divide powers by subtracting exponents. Then,

125 = b³

b = 5

(You can solve this, if you wish, by saying 5³ = b³ , so ( 5³)^1/3 = (b³)^1/3 or 5 = b

Then, as the function goes through (2, 75)

75 = a * 5²

75 = 25a

a = 3

y = 3 * 5^x

Answer 2

y=Ae^kx

ln y = ln a + kxinate k

ln (3/5) = ln A - k

ln 75 = ln A + 2k

This is a pair of simultaneous linear equations with unknowns ln A and k.

It solves pretty easily by elimination...start by subtracting the first equation from the 2nd which eliminates ln A.

The solution goes more easily if you then multiply the first equation by 2 and then add; this will eliminate k.