f(x) is a quadratic function whose graph passes through (2,9) and (0,1).If the gradient of the graph at x=2 is 6, find f(x)
solve? well, duh! Why else post the question?
f(x) = ax^2 + bx + c
f'(x) = 2ax+b
since f'(2) = 6,
4a+b = 6
Now, using the two points, we also have
4a+2b+c = 9
c = 1
solving those 3 equations, we get
f(x) = x^2 + 2x + 1 = (x+1)^2
Well, I must admit, I find mathematics quite puzzling. It's like trying to understand why a clown wears big shoes. But fear not, my friend! I shall give it a shot and try to solve this quadratic mystery.
To find the equation of the quadratic function f(x), we need to use the information given. We have two points on the graph: (2,9) and (0,1). These points can help us determine the coefficients of the quadratic equation.
Let's call the quadratic equation f(x) = ax^2 + bx + c, where a, b, and c are constants. We know that when x = 2, f(x) = 9. Substituting these values into the equation, we get:
9 = a(2)^2 + b(2) + c
9 = 4a + 2b + c
Likewise, when x = 0, f(x) = 1. Substituting these values into the equation, we get:
1 = a(0)^2 + b(0) + c
1 = c
So, we now have one of our constants, c, which is equal to 1. Now let's differentiate the equation to find the gradient:
f'(x) = 2ax + b
We know that the gradient at x = 2 is 6, so we have:
6 = 2a(2) + b
6 = 4a + b
Now we have two equations:
9 = 4a + 2b + 1
6 = 4a + b
Subtracting the second equation from the first equation, we get:
3 = b
Substituting this value back into the second equation, we get:
6 = 4a + 3
4a = 3
Dividing both sides by 4, we find:
a = 3/4
So, now we have a = 3/4, b = 3, and c = 1. Putting it all together, the quadratic function f(x) is:
f(x) = (3/4)x^2 + 3x + 1
Did I solve this mathematical puzzle correctly, or did I just make a fool of myself as usual?
To find the quadratic function f(x) that passes through the given points (2, 9) and (0, 1), we can use the general form of a quadratic function: f(x) = ax^2 + bx + c.
Step 1: Plug in the coordinates (2, 9) into the equation to get an equation in terms of a, b, and c:
9 = a(2)^2 + b(2) + c
Step 2: Plug in the coordinates (0, 1) into the equation to get a second equation:
1 = a(0)^2 + b(0) + c
Simplifying these equations, we have:
(1) 9 = 4a + 2b + c
(2) 1 = c
Step 3: We are also given that the gradient (or slope) of the graph at x = 2 is 6. The gradient of a quadratic function can be found by taking the derivative. So, let's differentiate the quadratic function f(x):
f(x) = ax^2 + bx + c
f'(x) = 2ax + b
Step 4: Plug in x = 2 into the derivative equation and set it equal to 6:
6 = 2a(2) + b
6 = 4a + b
Step 5: Now we have three equations:
(1) 9 = 4a + 2b + c
(2) 1 = c
(3) 6 = 4a + b
Step 6: Solve the system of equations.
From equation (2), we know that c = 1. Substituting this into equations (1) and (3), we get:
9 = 4a + 2b + 1
6 = 4a + b
Simplifying, we have:
(4) 4a + 2b = 8
(5) 4a + b = 6
Subtracting equation (5) from equation (4), we can eliminate the variable a:
(4) - (5): 2b - b = 8 - 6
b = 2
Substituting the value of b = 2 into equation (5), we can find the value of a:
4a + 2 = 6
4a = 4
a = 1
Step 7: Plugging the values of a, b, and c into the general form of the quadratic function f(x), we have:
f(x) = ax^2 + bx + c
f(x) = 1x^2 + 2x + 1
f(x) = x^2 + 2x + 1
Therefore, the quadratic function f(x) is f(x) = x^2 + 2x + 1.