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Solving quadratic

  • Thread starter Ewan_C
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  • #1
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[SOLVED] Solving quadratic

1. Homework Statement

1a) Solve f(x) = x^2- 4x+m in the form f(x) = (x-a)^2+ b
1b) What is the smallest value f(x) can have?


3. The Attempt at a Solution

1a) Seems simple enough. I set f(x) to 0 and used the completing the square method to solve. Ended up with f(x)=(x-2)^2+ m-4.

I don’t know how to approach 1b) though. I’m assuming I’ve made a mistake somewhere – there is no smallest value f(x) can have without knowing the value of m. Is there a way to find m that I haven't picked up on, or would the answer just be m - 4?

Thanks for any advice
 
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Answers and Replies

  • #2
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1. Homework Statement

1a) Solve f(x) = x^2- 4x+m in the form f(x) = (x-a)^2+ b
1b) What is the smallest value f(x) can have?


3. The Attempt at a Solution

1a) Seems simple enough. I set f(x) to 0 and used the completing the square method to solve. Ended up with f(x)=(x-2)^2+ m-4.

I don’t know how to approach 1b) though. I’m assuming I’ve made a mistake somewhere – there is no smallest value f(x) can have without knowing the value of m. Is there a way to find m that I haven't picked up on, or would the answer just be m - 4?

Thanks for any advice
Looks fine to me.
 
  • #3
HallsofIvy
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Homework Helper
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1. Homework Statement

1a) Solve f(x) = x^2- 4x+m in the form f(x) = (x-a)^2+ b
1b) What is the smallest value f(x) can have?


3. The Attempt at a Solution

1a) Seems simple enough. I set f(x) to 0 and used the completing the square method to solve. Ended up with f(x)=(x-2)^2+ m-4.

I don’t know how to approach 1b) though. I’m assuming I’ve made a mistake somewhere – there is no smallest value f(x) can have without knowing the value of m. Is there a way to find m that I haven't picked up on, or would the answer just be m - 4?

Thanks for any advice
No, what you have given is fine. No matter what x is (x- 2)2 can never be lower than 0 so f(x)= (x-2)2 + m- 4 can never be lower than m- 4. Since you are not given a specific value of m, that is all you can do.
 
  • #4
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Okay, thanks for that
 

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