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Zib
06-19-2011, 03:59 PM
The question asks me to find the equation of the quadratic function in standard form that has zeros 1+the square root of 11 and 1-the square root of 11 and that contains the point (4, -6)
Can anyone tell me how to set this up to begin with?

JeffM
06-19-2011, 05:07 PM
The question asks me to find the equation of the quadratic function in standard form that has zeros 1+the square root of 11 and 1-the square root of 11 and that contains the point (4, -6)
Can anyone tell me how to set this up to begin with?
Suppose u and v are zeroes of f(x).

Then f(x) = a(x - u)(x - v). (The proof of that IMPORTANT relationship is messy).

Now can you solve the problem?

Zib
06-19-2011, 05:44 PM
Thank you, but what I'm having trouble with is how 1+the square root of 11 and 1-the square root of 11 could be fit into that form. Can you help me please?

JeffM
06-19-2011, 05:59 PM
Thank you, but what I'm having trouble with is how 1+the square root of 11 and 1-the square root of 11 could be fit into that form. Can you help me please?
They are just numbers. Substitute them into a(x - u)(x - v) and do the algebra.

a(x - u)(x - v) = ax[sup:122sqf83]2[/sup:122sqf83] - ax(u + v) + auv. Original post neglected to insert the x into ax(u + v).
u = 1 + 11[sup:122sqf83]1/2[/sup:122sqf83].
v = 1 - 11[sup:122sqf83]1/2[/sup:122sqf83].

So u + v = 2.
And uv = 1 - 11 = 10.

So put the quadratic into standard form. The only parameter left is in a. So what value of a results in a quadratic that goes through the stipulated point?

Zib
06-19-2011, 06:36 PM
3! Thank you!
Except, it only works when I expand (x - u)(x - v) = ax2 - ax(u + v) + auv? Did you just miss a key, or does that expansion only work by coincidence?

mmm4444bot
06-19-2011, 11:37 PM
I expand (x - u)(x - v) = ax2 - ax(u + v) + auv? Did you just miss a key

Yes -- Jeff intended to type -ax(u + v)

Looks like you got: f(x) = 3x^2 - 6x - 30

Good work. 8-)

PS: Note the caret symbol in red? That's how we text exponent notation. (It's a shift 6, on most keyboards.)

And, we can text "square root of 11" as sqrt(11).

JeffM
06-20-2011, 12:11 AM
Yes I miskeyed. Sorry for any confusion it may have caused.

Zib
06-20-2011, 10:56 PM
Thank you both :D