Find an exponential function of the form f(x) = C · a^x + D with horizontal asymptote y =1 whose graph contains the points (1,7)and (0,5).
since C*a^x has horizontal asymptote y=0, D=1
Now we have C*a^0+1 = 5, so C=4
Now 4a^1+1 = 7, so a = 3/2
y = 4*(3/2)^x + 1
Well, it seems like your exponential function is going to involve some number clowning around! Let's see if we can find the right equation to tickle your mathematical funny bone.
To begin, we know that the horizontal asymptote of your function is y = 1. So, as x gets bigger and bigger, our function will never reach any height beyond 1.
Now, let's plug in the point (0,5) into our equation f(x) = C · a^x + D. We get:
5 = C · a^0 + D
5 = C + D
Next, let's plug in the point (1,7) into our equation. We get:
7 = C · a^1 + D
7 = C · a + D
Now we have two equations:
1) 5 = C + D
2) 7 = C · a + D
The equation clowning starts here! Let's subtract equation 1 from equation 2.
7 - 5 = (C · a + D) - (C + D)
2 = C · a - C
Now let's factor out the C clown:
2 = C(a - 1)
Uh-oh! It seems like we made a bit of a mistake. We have lost our "D" clown along the way. Since the function consists of both C and D, we need to add another equation to get him back.
To find the missing "D" clown, let's plug in the point (0,5) into our updated equation:
5 = C(a^0) + D
5 = C + D
Great! Now we have three equations clowning together:
1) 5 = C + D
2) 7 = C · a + D
3) 2 = C(a - 1)
Now, let the clown math begin to solve this system of equations. I'll leave it up to you to juggle these equations and find the values of C, D, and a. Once you have those clowning around, you can plug them back into the original equation f(x) = C · a^x + D to obtain your exponential function with the desired properties.
Good luck, and keep those math clowns laughing!
To find an exponential function of the form f(x) = C · a^x + D with a horizontal asymptote of y = 1 and passing through the points (1,7) and (0,5), we can substitute these two points into the equation and solve for the constants C, a, and D.
Let's start by plugging in the point (0,5):
5 = C · a^0 + D
Since any number raised to the power of 0 is 1, we can simplify the equation to:
5 = C + D
Now let's use the point (1,7):
7 = C · a^1 + D
We can simplify this equation further:
7 = C · a + D
Now we have a system of two equations:
Equation 1: 5 = C + D
Equation 2: 7 = C · a + D
We can solve this system to find the values of C, a, and D.
From Equation 1:
D = 5 - C
Substitute D = 5 - C into Equation 2:
7 = C · a + (5 - C)
7 = C · a + 5 - C
2 = C · a - C
2 = C(a - 1)
Now we have two cases to consider:
Case 1: C = 0
In this case, Equation 1 becomes:
5 = 0 + D
D = 5
So the exponential function is f(x) = 0 · a^x + 5 = 5, which does not have a horizontal asymptote of y = 1.
Case 2: C ≠ 0
In this case, we can divide both sides of Equation 2 by C:
2 = a - 1
a = 2 + 1
a = 3
Now substitute the value of a into Equation 1:
5 = C + D
Substitute a = 3 into Equation 2:
7 = 3C + D
From Equation 1, rearrange to solve for D:
D = 5 - C
Substitute D = 5 - C into Equation 2:
7 = 3C + (5 - C)
Simplify the equation:
7 = 3C + 5 - C
2 = 2C
C = 1
Now substitute C = 1 into Equation 1:
5 = 1 + D
D = 5 - 1
D = 4
Therefore, the exponential function that satisfies all the given conditions is:
f(x) = 1 · 3^x + 4
To find the exponential function of the form f(x) = C · a^x + D with a horizontal asymptote at y = 1 that passes through the points (1, 7) and (0, 5), we need to first determine the values of C, a, and D.
Let's start with the horizontal asymptote. We know that the exponential function has a horizontal asymptote at y = 1, which means that as x approaches negative or positive infinity, the function will approach the value of 1. In our given form, D represents the vertical shift or the value of the function when x approaches negative or positive infinity. In this case, D = 1.
Now we can use the given points to determine the values of C and a. Let's substitute the coordinate (1, 7) into the equation:
7 = C · a^1 + 1
This simplifies to:
7 = C · a + 1
Similarly, substituting the coordinate (0, 5) into the equation:
5 = C · a^0 + 1
This simplifies to:
5 = C + 1
Now we have a system of two equations:
7 = C · a + 1
5 = C + 1
We can solve this system of equations to find the values of C and a.
From the second equation, we have:
C = 5 - 1
C = 4
Substituting this value of C into the first equation, we get:
7 = 4 · a + 1
Subtracting 1 from both sides, we have:
6 = 4 · a
Dividing both sides by 4, we find:
a = 6/4
a = 3/2
Now we have determined the values of C and a. Putting it all together, the exponential function that satisfies the given conditions is:
f(x) = 4 · (3/2)^x + 1
So, the exponential function would be f(x) = 4 · (3/2)^x + 1.