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#1 2005-11-30 04:41:59

RickyOswaldIOW
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More Quadratics

I'm given the function f(x)=-(x-2)^2+9 and asked for the maximum value (9) and to sketch the graph of y=f(x) showing where the curve crosses the axis.  I'm not sure of the standard way to work backwards but I worked back to y = -x^2 + 4x + 5 and tried to factorise it to get the points where it crosses the x-axis.
Factorising gives me -(x-5)(x-1) [[crosses at (5,0) and (1,0)]] which ofc does not fit in with the sketch.  I'm begining to think that maybe I should factorise down to numercial order thus -(x-1)(x-5) and since the first bracket is preceeded by a - the sign change is canceled out giving (-1,0)(5,0)?
Maybe if someone could post the standard rule to work this out, it would be tres handy

Aloha Nui means Goodbye.

#2 2005-11-30 06:01:40

mathsyperson
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Re: More Quadratics

Factorising -x² + 4x + 5 gives you (x+1)(5-x), so the roots would be -1 and 5.
You had the right answer, but wrong reasoning.

Why did the vector cross the road?
It wanted to be normal.

#3 2005-11-30 06:06:22

RickyOswaldIOW
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Re: More Quadratics

I know the answer was right, I looked it up
My question (which I realise is not that clear) was,  "How do I work it out?"

Aloha Nui means Goodbye.

#4 2005-11-30 08:34:24

irspow
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Re: More Quadratics

I would personally use the quadratic equation for this problem:

-x^2 + 4x + 5

The quadratic equation would yield the same thing without trying different combinations.

[ -b +/- (b^2 - 4ac)^(1/2)] / 2a   =  [-4 +/- (16 + 20)^(1/2)] / -2, so x = 5 and -1

Maybe I am biased, but this method always seemed easier than figuring out values by trial and error.

Your question though was specifically about the "textbook" method of factoring.  I would describe it as follows.

You want your answer to be like: (ax + b)(cx + d).  This equals (ac)x^2 + (ad + bc)x + (bd).
So you are really concerned with the coefficients in the original equation.

In this case: ac = -1,  ad + bx = 4, and bd = 5.  b and d are the second terms, so just ask yourself what two numbers multiply to produce 5.  The simple answer leaves only one and five.    (and if you later find that you need fractional coefficients, God help you.)

(    + 1)(     + 5).......they both must be postitive otherwise the constant would be negative.

As ac = -1 and a and c represent the first terms just plug in a one into both brackets. Remember that one of these terms must be negative!

(1x  +1)(1x  +5)

The signs within the brackets come from the ad + bc term that produces the x^1 term in your original equation.  So ad = 5 and bc = 1 and these two answers must add up to 4.  The only way that a 5 and a 1 can equal 4 is if the 1 is negative.  Therefore the bc term must be negative and b is already a positive one so c must be negative.

(x+1)(-x+5)

Remember the (ac)x^2 + (ad + bc)x + bd form of the equation to keep all of this straight in the harder factoralizations.

Now you see why I almost always use the quadratic equation to find the zeros of a 2nd degree equation.  Just plug in the coefficients and your done.

#5 2005-11-30 09:09:06

RickyOswaldIOW
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Re: More Quadratics

Wow thanks, I'll make my answer a little more specific next time.

Again, Thanks

Aloha Nui means Goodbye.

#6 2005-12-01 12:50:29

RickyOswaldIOW
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Re: More Quadratics

I've been using the quadratic formula to solve equations on which I first complete the square.  I notice you are using the same formula here but without the sqrt?

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#7 2005-12-02 08:18:21

irspow
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Re: More Quadratics

If you look inside the brackets of the numerator the sum within them is raised to the one-half power which is the same thing as taking the square root but just different notation.

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