Determine the values of a and b so that the function f(x)=x^2+ax+b has a minimum at (3, 9).
f(x)=x^2+ax+b
f ' (x) = 2x + a
but f ' (3) = 0
2(3) + a = 0
a = -6
so f(x) = x^2 - 6x + b
we also know that f(3) = 9
3^2 - 6(3) + b = 9
b = 18
Well, at least we know that this function is not clowning around with a maximum! Let's put on our detective hats and solve the mystery.
To find the minimum of a quadratic function, we need to use some detective work and remember a key clue: at the minimum point, the derivative of the function is equal to zero.
Let's take the derivative of f(x) with respect to x:
f'(x) = 2x + a
Since the minimum point is (3, 9), we know that when x = 3, f'(x) = 0:
0 = 2(3) + a
0 = 6 + a
a = -6
Now that we've unraveled the mystery of a, let's find b. We substitute the values of a and (3, 9) into our original function:
9 = 3^2 + (-6)(3) + b
9 = 9 - 18 + b
9 = -9 + b
b = 18
So the values of a and b that satisfy the conditions are a = -6 and b = 18. Well, it seems like the quadratic function didn't have any funny business this time. Case solved!
To determine the values of a and b such that the function f(x) = x^2 + ax + b has a minimum at (3, 9), we can use the fact that the vertex of a parabola represents the minimum point.
The vertex form of a quadratic function is given by f(x) = a(x - h)^2 + k, where (h, k) represents the vertex of the parabola.
In this case, the vertex is given as (3, 9), so we have the values of h=3 and k=9. Substituting these values in the vertex form of the function, we have:
f(x) = a(x - 3)^2 + 9
Now we can equate this equation to the given function f(x) = x^2 + ax + b:
x^2 + ax + b = a(x - 3)^2 + 9
Expanding the right side of the equation:
x^2 + ax + b = ax^2 - 6ax + 9a + 9
Now, let's equate the coefficients of x^2, x, and the constant term on both sides of the equation:
1 = a (coefficient of x^2)
a = -6a (coefficient of x)
b = 9a + 9 (constant terms)
From the equation a = -6a, we can solve for a:
7a = 0
a = 0
Now substituting the value of a into the equation for b, we have:
b = 9(0) + 9
b = 9
Therefore, the values of a and b such that the function f(x) = x^2 + ax + b has a minimum at (3, 9) are a = 0 and b = 9.
To find the values of `a` and `b` such that the function `f(x) = x^2 + ax + b` has a minimum at the point (3, 9), we can utilize the fact that the vertex of a parabola represents its minimum or maximum point.
The vertex form of a quadratic function is given by `f(x) = a(x - h)^2 + k`, where `(h, k)` represents the coordinates of the vertex.
In this case, we know that the vertex is (3, 9), so we can substitute these values into the vertex form equation:
9 = a(3 - h)^2 + k
Since the vertex is (3, 9), we have h = 3 and k = 9:
9 = a(3 - 3)^2 + 9
Simplifying this equation further:
9 = a(0)^2 + 9
9 = a(0) + 9
9 = 0 + 9
Thus, we can conclude that a can be any real number since it doesn't affect the minimum value of the parabola. There are infinitely many values of `a` that can be chosen.
The value of `b` can be determined by substituting the x-coordinate of the vertex (3) and the y-coordinate of the vertex (9) into the original function:
f(3) = 3^2 + 3a + b = 9
9 + 3a + b = 9
3a + b = 0
Here we have an equation relating `a` and `b`. As mentioned earlier, `a` can be any real number, so we can solve this equation for `b`:
b = -3a
Therefore, the values of `a` and `b` that satisfy the requirement that the minimum of the function `f(x) = x^2 + ax + b` occurs at (3, 9) are given by `a` can be any real number, and `b` is equal to `-3a`.