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Find the exponential function of the form y=ax^b whose graph passes through the points (-3, 1/27) and (0,1).

I don't know where to begin with this problem??

• Advanced Algebra - typo? - ,

(0,1) means when x=0, y=1.
When x=0,
y = a*0^b = 0 except when b=0, when y is undefined.
So y cannot equal 1 when x=0.

If this is the case, then
at (0,1),
1=ab^0=a*1=a
so y=b^x
at (-3,1/27)
1/27 = b^(-3)
27 = b^3;
3^3=b^3
b=3

So what do i put as the answer? is 3 the esponential function then?

"Please check if there is a typo,..."

When you see what the question actually asks, then you can decide which way to go.
Especially in exams, you need to double-check what you've copied the question correctly before you start.

Im sorry yes there was a typo, you correctly fixed it. the b should be an x so it should look like this.
y=ab^x

So the second solution would be appropriate. BUT... do make sure you understand how the solution was obtained, promise?

I do! thank you so much!!

• Advanced Algebra :) - ,

You're welcome!