# Quadratic function solution method optimizing

• YoungPhysicist
In summary: Since the minimum value of a parabola is at its vertex, and this parabola has a minimum value of 6, the vertex must be at (d,6).Using the vertex form, we can see that the value of d is equal to the x-coordinate of the vertex.Therefore, d = 0.Since d = 0, b = -2d = -2(0) = 0.Therefore, b = 0.In summary, the quadratic function ƒ(x) = x2-2bx+7 will have a minimum value of 6 when b = 0.
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?

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

scottdave and YoungPhysicist
verty said:
The minimum or maximum value of a parabola is ##f({-b \over 2a})## where ##f(x) = ax^2 + bx + c##.
Thanks!

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
scottdave said:
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.

Young physicist said:

## 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?

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)

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## 1. What is a quadratic function?

A quadratic function is a mathematical expression that can be written in the form of f(x) = ax^2 + bx + c, where a, b, and c are constants and x is the variable. It is a type of polynomial function that can be graphed as a parabola.

## 2. What is the solution method for quadratic functions?

The solution method for quadratic functions is the process of finding the values of x that make the function equal to zero. This can be done through various methods, such as factoring, completing the square, or using the quadratic formula.

## 3. How do you optimize a quadratic function?

To optimize a quadratic function, you need to find the maximum or minimum value of the function. This can be done by finding the vertex of the parabola, which is the point where the function reaches its highest or lowest point. This can also be done by using calculus and finding the derivative of the function.

## 4. What is the difference between finding the solution and optimizing a quadratic function?

Finding the solution of a quadratic function means finding the values of x that make the function equal to zero. This can be done through various methods, as mentioned before. Optimizing a quadratic function, on the other hand, means finding the maximum or minimum value of the function, which can also be done through various methods.

## 5. Can a quadratic function have more than one solution?

Yes, a quadratic function can have two solutions, one solution, or no solutions at all. The number of solutions depends on the discriminant of the function, which is b^2 - 4ac. If the discriminant is positive, the function will have two real solutions. If it is zero, the function will have one real solution. If it is negative, the function will have no real solutions.

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