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Quadratic function solution method optimizing

  • #1

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:

Answers and Replies

  • #2
verty
Homework Helper
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The minimum or maximum value of a parabola is ##f({-b \over 2a})## where ##f(x) = ax^2 + bx + c##.
 
  • #3
The minimum or maximum value of a parabola is ##f({-b \over 2a})## where ##f(x) = ax^2 + bx + c##.
Thanks!
 
  • #4
scottdave
Science Advisor
Homework Helper
Insights Author
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701
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.
 
  • #5
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.:smile:
 
  • #6
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
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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.
 
  • #7
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)

65F26413-34DC-4441-A85E-E4070C507F83.jpeg
 

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