b^0 = 1, so a=3
Now you know that 3b^2 = 12
I expect you can find b now ...
The sketch shows a curve with equation y=ab^x where a and b are constants and b>0.
The curve passes through the points (0,3) and (2,12).
Calculate the value of a and b.
Thank you.
2 answers
sub in the two points to set up two equations
for (0,3)
3 = a b^0, but b^0 = 1
so a = 3
for (2,12)
12 = 3 b^2
4 = b^2
b = ± 2, but we need b> 0
a = 3 , b = 2 ---> y = 3 (2)^x
for (0,3)
3 = a b^0, but b^0 = 1
so a = 3
for (2,12)
12 = 3 b^2
4 = b^2
b = ± 2, but we need b> 0
a = 3 , b = 2 ---> y = 3 (2)^x