Question

What is the exponential function for the graph passing through (-2,1) and (-1,2)

Answer 1

To find the exponential function for the graph passing through (-2,1) and (-1,2), we need to use the general form of an exponential function, which is:

y = a * b^x

where:

a is the initial value or the y-intercept of the function

b is the base of the exponential function

x is the variable or the exponent

To determine the values of a and b, we need to use the two given points and solve for the corresponding equations. Substituting the first point (-2,1), we get:

1 = a * b^(-2)

Substituting the second point (-1,2), we get:

2 = a * b^(-1)

Now, we can solve for a and b by eliminating one variable. Dividing the second equation by the first equation, we get:

2/1 = a * b^(-1) / (a * b^(-2))

2 = b

Substituting this value of b into the first equation, we get:

1 = a * 2^(-2)

a = 4

Therefore, the exponential function that passes through (-2,1) and (-1,2) is:

y = 4 * 2^x

or

f(x) = 4 * 2^x

Answer 2

An exponential function has the form `f(x) = ab^x`, where `a` and `b` are constants. To find the exponential function that passes through the points `(-2,1)` and `(-1,2)`, we can use these points to set up a system of equations to solve for the values of `a` and `b`.

Substituting the coordinates of the first point into the equation for an exponential function gives us:

`1 = ab^(-2)`

Substituting the coordinates of the second point into the equation for an exponential function gives us:

`2 = ab^(-1)`

Dividing these two equations gives us:

`(2)/(1) = (ab^(-1))/(ab^(-2))`

`2 = b`

Now that we know that `b=2`, we can substitute this value back into either equation to solve for `a`. Substituting into the first equation gives us:

`1 = a(2)^(-2)`

`1 = a(1/4)`

`a = 4`

So, the exponential function that passes through the points `(-2, 1)` and `(-1, 2)` is given by:

`f(x) = 4 * 2^x`.