I don't understand how to do this problem. Please help!
Find an exponential function of the form f(x) = bax that has a y-intercept of 2 and passes through the point (2, 98). For example, if the equation is f(x) = 2(7)x respond by entering the answer as 2(7)^x
I think you meant:
y = b(a^x)
you have 2 points, (0,2) and (2,98) which lie on this graph
sub in (0,2)
2 = b(a^0) ---> b = 2
so y = 2(a^x), now sub in (2,98)
98 = 2(a^2)
49 = a^2
a = ± 7
so your equation is y = 2(7)^x
(strange that they would give you the actual answer as a hint or example)
To find the exponential function that satisfies the given conditions, we'll use the information provided.
The y-intercept of the function is given as 2, which means that when x = 0, the value of f(x) is 2. This gives us the point (0, 2) on the graph.
We are also given that the function passes through the point (2, 98). So when x = 2, the value of f(x) is 98.
Now, let's use the general form of an exponential equation: f(x) = b * a^x, where b represents the initial value or y-intercept, and a is the base.
Substituting the y-intercept values into the equation, we have:
f(0) = 2, which gives 2 = b * a^0, and since a^0 = 1, we have 2 = b * 1, so b = 2.
So we now have the equation f(x) = 2 * a^x.
Using the second point given, (2, 98), we can substitute x = 2 and f(x) = 98 into the equation and solve for a:
98 = 2 * a^2.
Dividing both sides of the equation by 2 gives:
49 = a^2.
Taking the square root of both sides, we get:
a = ±√49.
So there are two possible values for a: a = 7 and a = -7.
Therefore, the equation that satisfies the given conditions is:
f(x) = 2 * 7^x or f(x) = 2 * (-7)^x.