Quadratic function solution method optimizing

[YoungPhysicist]

Homework Statement

For which values of b will the quadratic function ƒ(x) = x²-2bx+7 have a minimum value of 6?

Homework Equations

y = ax²+bx+c
y = a(x-h)²+k

b(first one) = -2ah(second one)
c(first one) = ah²+k(second one)

The Attempt at a Solution

Well, I actually already have the answer to the question: b = ±1, but I believe there must be a better method to this.

My poor method goes like this:

Since the parabola has a "minimum", it must open upwards,with a vertex's y position as 6, which makes the function I am dealing with looks like these two forms:

• y = x²-2bx+7
• y = (x-h)²+6

which the second one can be turned into y = x²-2hx+h²+6
∴ h²+6 = 7(c)
∴h = ±1

then because -2b = -2h,and h = ±1, b = ±1.

Is there a simpler way of doing this?

 

[verty]

The minimum or maximum value of a parabola is $f({-b \over 2a})$ where $f(x) = ax^2 + bx + c$.

 

[YoungPhysicist]

The minimum or maximum value of a parabola is $f({-b \over 2a})$ where $f(x) = ax^2 + bx + c$.

Thanks!

 

[scottdave]

With a minimum value of 6, this parabola doesn't have real roots, but if we took a shifted parabola down by 6, the vertex would just touch the x axis. Thia happens when the discriminant of the quadratic formula is zero. If you subtract 6 then solve for the discriminant equal zero.

 

[YoungPhysicist]

With a minimum value of 6, this parabola doesn't have real roots, but if we took a shifted parabola down by 6, the vertex would just touch the x axis. Thia happens when the discriminant of the quadratic formula is zero. If you subtract 6 then solve for the discriminant equal zero.

Thanks,too. Yet another great method.

 

[Ray Vickson]

Homework Statement

For which values of b will the quadratic function ƒ(x) = x²-2bx+7 have a minimum value of 6?

Homework Equations

y = ax²+bx+c
y = a(x-h)²+k

b(first one) = -2ah(second one)
c(first one) = ah²+k(second one)

The Attempt at a Solution

Well, I actually already have the answer to the question: b = ±1, but I believe there must be a better method to this.

My poor method goes like this:

Since the parabola has a "minimum", it must open upwards,with a vertex's y position as 6, which makes the function I am dealing with looks like these two forms:

• y = x²-2bx+7
• y = (x-h)²+6

which the second one can be turned into y = x²-2hx+h²+6
∴ h²+6 = 7(c)
∴h = ±1

then because -2b = -2h,and h = ±1, b = ±1.

Is there a simpler way of doing this?

Your method is about as simple as it gets.

 

[YoungPhysicist]

Thanks everyone. Problem solved. Here is my improved method based of @verty 's post:

ƒ(x) = x²-2dx+7(easier to distinguish from the b in y = ax²+bx+c)