# 1. What is the y-intercept of the graph of `y=a^(x-2)` y-intercept: answer is `a^(-1)` Please tell me how to get there!

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To solve an exponential graph, substitute the values such that if x= 0 (which indicates the y-intercept):

`y=a^(x-2)` becomes : `y=a^(0-2)`

`therefore y=a^-2` Therefore(x;y): `(0;a^-2)`

However this is not the answer required in terms of the question.

In order to get an answer of `y=a^-1` we need to substitute x=1:

`therefore y= a^(1-2)` Therefore (x;y): `(1;a^-1)`

This is NOT the y intercept as x=0 at that point.

**Ans:**

**`(0; a^-2)` when x=0 or `(1;a^-1)` when y=`a^-1` **

`y=a^(x-2)`

y-intercept of this graph depends on **a. **

If `a>0`

substitute x=0 in given equation

`y=a^(0-2)=1/a^2` is y intercept.

`if a<0`

substitute x=0 in given equation

`y=(-a)^(0-2)=1/a^2`

If a=0 then

`y=0^(x-2)=0AAx ` except x=2 because `y=0^0` is indeterminant form.

intercept is (1/4)

`y=(-2)^(x-2)`

We unable to plot graph.

For the graph of `y=a^(x-2)` , we can find the y-intercepts by substituting x=0. That is

`y=a^(0-2) =a^(-2)=1/a^2`

` `

Actually when a<0 and x takes fractional values the fuction will not remain in R. So we can not take a<0 for all real values of x.