find the parabola whose minimum is at (7,6) rather than the point given in the book.
The parabola's equation is y = x2 + ax + b, where
a = and
b =
(dy/dx)=2x+a
Funcktion have extreme value where is
(dy/dx)=0
2x+a=0 ,2x=-a, Divided with 2,
x=-(a/2)
For x=7 , 7=(-a/2) , Multipled with 2
7*2=-a , 14=-a , Multilpled with (-1)
a=-14
y=x^2+ax+b
x=7 y=6 a=-14
6=(-a/2)^2+a*(-a/2)+b
6=(14/2)^2-14*(-14/2)+b
6=7^12-14*7+b
6=49-98+b , 6=-49+b , b=6+49 , b=55
y=x^2+ax+b
y=x^2-14x+55
For x=7
y=7^2-14*7+55=49-98+55=6
Proof that is minimum.
Function have minimum if (dy/dx)=0
and (d^2y/dx^2)>0
If second derivate greater than zero.
Second derivate is derivate of first derivate.
In this case:
(d^2y/dx^2)=d(2x)/dx=2>0
That is minimum
To find the values of a and b when the minimum of the parabola is at (7,6), you can substitute the coordinates of the vertex into the equation for y.
Given that the minimum is at (7,6), the x-coordinate of the vertex is 7 and the y-coordinate is 6.
Substituting these values into the equation y = x^2 + ax + b:
6 = (7)^2 + 7a + b
Simplifying:
6 = 49 + 7a + b
Rearranging the equation:
7a + b = 6 - 49
7a + b = -43
Now you have one equation in two variables (a and b). The equation alone doesn't allow you to determine the specific values of a and b. There are infinitely many combinations of a and b that can satisfy this equation.
However, if you have additional information or conditions, you may be able to determine specific values for a and b.
To find the equation of the parabola with a minimum at (7,6) instead of the given point in the book, we need to find the values of a and b.
Given that the minimum point (h, k) of a parabola in the standard form y = ax^2 + bx + c satisfies the coordinates (h, k), we can substitute the values (7, 6) into the equation y = x^2 + ax + b to obtain:
6 = 7^2 + 7a + b
Simplifying this equation, we get:
6 = 49 + 7a + b
Rearranging the terms, we have:
7a + b = 6 - 49
7a + b = -43
Thus, we have one equation with two variables. We need one more equation to solve for both a and b.
One way to find the additional equation is to use the vertex form of a parabola. In the vertex form, the equation is written as y = a(x - h)^2 + k, where the vertex is given by (h, k).
Since we know the vertex is (7, 6), we can substitute these values into the vertex form equation to get:
y = a(x - 7)^2 + 6
Setting x = 7, we can find the y-coordinate of the vertex:
6 = a(7 - 7)^2 + 6
6 = a(0) + 6
6 = 6
This equation gives us a second equation, 0a = 0, which means any value of a will satisfy this equation.
Now we can solve the system of equations:
7a + b = -43
0a = 0
Since the second equation indicates that a can have any value, we can choose any arbitrary value for a. For simplicity, let's choose a = 1.
Substituting a = 1 into the first equation, we get:
7(1) + b = -43
7 + b = -43
b = -43 - 7
b = -50
Therefore, the parabola's equation with a minimum at (7,6) is:
y = x^2 + x - 50