Find the exponential equation whose graph passes through the points (-1,2) and (1,8) y =

Question

Find the exponential equation whose graph passes through the points (-1,2) and (1,8)

y =   

Answer 1

(-1,2) and (1,8)

y=a(b^x)

2=ab^-1

a=2b

8=ab^1 = ab

ab=8

2b(b) =8

b^2 = 4

b=+/-2

a =+/-4

y=4(2)^x

is the exponential equation you probably want

also the following equation fits the two given points

y =-4(-2)^x

^{but it only works for integer values of x. it's a discontinous curve graphically}

^{a negative number raised to a fractional exponent is undefined}

^{both are exponential equations, so the correct answer is both, but the instructor, textbook, or online answer is probably just the positive values of a and b}

Answer 2

Let us write an exponential equation in the form

y = A · B ^x, where B > 0

Since its graph passes through the point ( -1 , 2 ), we can substitute x = -1 and y = 2 and obtain

2 = A · B ^-1 = A / B

Similarly, since its graph passes through the point ( 1 , 8 ), we can substitute x = 1 and y = 8 and obtain

8 = A · B ¹ = A · B

Now we have a system of two equations with two unknowns A and B

A · B = 8

A / B = 2

There are many ways to solve it. The one I like is to simply divide the first equation by the second.

On the left side we will get (A · B) / (A / B) = B² and on the right side we will get 8 / 2 = 4

So B² = 4, therefore B = -2 or B = 2. But since B has to be positive, the only possibility is B = 2

Substituting B = 2 into the first equation, we get

A · 2 = 8 or A = 4

So the exponential equation we are looking for is

y = 4 · 2^x

The last step is to check our work

If x = -1, y = 4 · 2^x = 4 · 2^-1 = 4 / 2 = 2

If x = 1, y = 4 · 2^x = 4 · 2¹ = 4 · 2 = 8