A parabola, which is given by the equation y = x^(2)+ax+b, attains its minimum at (6,8). Find a and b.
a =
b =
The vertex is at (6,8) so the equation is
y = m(x-6)^2 + 8 = mx^2 - 12mx + 36m+8
m=1, so
a = -12
b = 44
Well, solving this problem is no laughing matter! Let's get down to business.
To find the values of a and b, we need to use the given information that the parabola attains its minimum at the point (6, 8).
First, let's substitute the coordinates of the minimum point into the equation:
8 = 6^2 + 6a + b
Simplifying this equation, we get:
8 = 36 + 6a + b
Now, we can use the fact that the parabola attains its minimum at this point to find the values of a and b.
The minimum point of a parabola occurs at x = -b/2a, and in this case, we know that x = 6.
Plugging in x = 6 into the equation, we get:
6 = -b/2a
Rearranging this equation, we have:
12a = -b
Now we have two equations:
8 = 36 + 6a + b
12a = -b
Solving these two equations simultaneously is quite a circus act! Let's solve them together. Subtracting the second equation from the first gives us:
8 = 36 + 6a + b
-12a = b
Substituting -12a for b in the first equation, we get:
8 = 36 + 6a - 12a
Simplifying this equation, we have:
8 = 36 - 6a
Rearranging, we get:
6a = 36 - 8
6a = 28
a = 28/6
a = 14/3
Now that we have found the value of a, we can substitute it back into the equation -12a = b to find b:
b = -12 * (14/3)
b = -168/3
b = -56
So, the values of a and b that satisfy the given conditions are:
a = 14/3
b = -56
I hope I was able to make this problem a little more fun for you!
To find the values of a and b in the equation y = x^2 + ax + b, given that the parabola attains its minimum at the point (6,8), we can use the vertex form of a parabola.
The vertex form of a parabola is given by y = a(x - h)^2 + k, where (h, k) is the vertex of the parabola.
In this case, we know that the minimum point of the parabola is (6,8). So the vertex coordinates are (h, k) = (6, 8).
Substituting this into the vertex form equation, we get:
8 = a(6 - 6)^2 + k
8 = a(0)^2 + k
8 = k
So we have k = 8.
Now, substitute the vertex coordinates (h, k) = (6, 8) and any other point on the parabola into the equation to solve for a:
8 = a(6 - 6)^2 + 8
8 = a(0) + 8
8 = 8
Therefore, a can be any value since it does not affect the equation of the parabola.
So, a is any real number, and b is 8.
To solve this problem, we will use the information given about the minimum point of the parabola.
The vertex form of a parabola is given by: y = a(x-h)^2 + k, where (h,k) represents the coordinates of the vertex. In this case, we know that the minimum point is at (6,8), so the vertex form can be written as: y = a(x-6)^2 + 8.
We also have the equation of the parabola given as y = x^2 + ax + b. By comparing this equation with the vertex form, we can see that a=1 and b=0.
Let's substitute these values into the equation y = a(x-6)^2 + 8:
y = 1(x-6)^2 + 8
y = (x-6)^2 + 8
Now we can expand the square:
y = x^2 - 12x + 36 + 8
y = x^2 - 12x + 44
By comparing this equation with the given equation y = x^2 + ax + b, we can conclude that a = -12 and b = 44.
Therefore, we have:
a = -12
b = 44
y = x^2 + ax + b
(6,8) lies on it, so 8 = 36 + 6a + b
6a + b = -28
dy/dx = 2x + a
when x = 6, the dy/dx = 0
2(6) + a = 0
a = -12