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chavi.rg
Students
 
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Joined: Sun Jan 20, 2013 11:35 am
 

Algebra Medium Practice Set Q9

by chavi.rg Tue Apr 09, 2013 10:36 am

Hi,

Question 9 in this gives you a function f(x) = x^2 +4x -5 and asks for its min value. I've never seen a question regarding max/min values for a function and I wasn't sure how to do it.

How do you find the min value of a function? Do we use -b/2a as in the vertex of a Parabola?

Thanks,
Chavi
tommywallach
Manhattan Prep Staff
 
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Joined: Thu Mar 31, 2011 11:18 am
 

Re: Algebra Medium Practice Set Q9

by tommywallach Sun Apr 14, 2013 11:25 pm

Hey Chavi,

The best way to find a minimum is just to test values. It'll never take long.

f(x) = x^2 + 4x - 5

We can already see that positive numbers aren't going to get us to a minimum, so let's start searching downwards.

x = 0, f(x) = 0 + 0 - 5 = -5
x = -1, f(x) = 1 - 4 - 5 = -8
x = -2, f(x) = 4 - 8 - 5 = -9
x = -3, f(x) = 9 - 12 - 5 = -8

Looks like we found the minimum! It's x = -2, f(x) = -9. Tada!

-t

P.S. This method is discussed in the book. There's no need for graphing/a graphing calculator ever, but it is definitely reasonable for them to ask for the minimum value of a function (it'll always be an integer, though, or else it would be way too hard to find).