Find an exponential function of the form
f(x) = ba^x that has the given y-intercept and passes through the point P.
y-intercept 4; P(2, 1)
a^x always passes through (0,1), so
f(x) = 4a^x
1 = 4a^2
a = 1/2
f(x) = 4 * (1/2)^x
Why did the exponential function cross the y-axis? Because it couldn't resist starting at 4...get it, 4...y-intercept...okay, maybe it's not that funny. But let me calculate that for you!
To find the exponential function that passes through the point P(2,1) and has a y-intercept of 4, we can substitute the values into the equation f(x) = ba^x and solve for the missing values.
Since the y-intercept occurs at x = 0, we have f(0) = 4, which gives us the equation:
4 = ba^0
Simplifying, we have:
4 = b
Now, let's substitute the point P(2,1) into the equation:
1 = 4a^2
Dividing both sides by 4, we get:
1/4 = a^2
Taking the square root of both sides, we have:
a = ±√(1/4)
Therefore, there are two possible exponential functions:
f(x) = 4(√(1/4))^x
and
f(x) = 4(-√(1/4))^x
Hope that helps, and I apologize if my humor didn't quite make you laugh!
To find an exponential function that satisfies the given conditions, we need to solve for the values of b and a.
Step 1: Determine the y-intercept.
The y-intercept is given as 4, which means the function passes through the point (0, 4).
Step 2: Substitute the y-intercept into the equation.
Using the point (0, 4), we can substitute x = 0 and y = 4 into the equation f(x) = ba^x:
4 = ba^0
Since any number raised to the power of zero equals 1, we can simplify the equation to:
4 = b * 1
b = 4
So, we have found the value of b as 4.
Step 3: Determine the value of a using the point P(2, 1).
Using P(2, 1), we can substitute x = 2 and y = 1 into the equation f(x) = ba^x:
1 = 4a^2
Divide both sides of the equation by 4:
1/4 = a^2
Take the square root of both sides:
±√(1/4) = a
±(1/2) = a
So, we have found the potential values of a as ±(1/2).
Therefore, one possible exponential function that satisfies the given conditions is:
f(x) = 4(1/2)^x
To find an exponential function of the form f(x) = ba^x that has the given y-intercept and passes through the point P, we can use the y-intercept and the point P to form equations and solve for the values of b and a.
First, let's consider the y-intercept. The y-intercept is the point on the graph of the exponential function where x = 0. From the given information, we know that the y-intercept is (0, 4). This means that when x is 0, the value of f(x) is 4. We can substitute these values into the exponential function equation:
f(x) = ba^x
When x = 0, f(x) = 4:
4 = ba^0
4 = b(1)
4 = b
So, we have found that the value of b is 4.
Now, let's consider the point P(2, 1) which lies on the graph of the exponential function. We can substitute the x and f(x) values into the equation and solve for a:
f(x) = ba^x
When x = 2, f(x) = 1:
1 = 4a^2
Next, we can solve this equation for a. Divide both sides of the equation by 4:
1/4 = a^2
To eliminate the square on the right side of the equation, we take the square root of both sides:
√(1/4) = a
√(1/4) = 1/2
So, we have found that the value of a is 1/2.
Now that we have determined the values of b and a, we can substitute them back into the equation to get the exponential function:
f(x) = ba^x
f(x) = 4(1/2)^x
Therefore, the exponential function of the form f(x) = ba^x that has the y-intercept (0, 4) and passes through the point P(2, 1) is f(x) = 4(1/2)^x.