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Probably easy quadratics question


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#1 Dysfunction

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Posted 06 October 2009 - 11:39 PM

f(x)  =  -3x^2  +  36x  -  150
Label X&Y intercepts, turning point and sketch the graph.

Y Intercept = -150

X  Intercepts have got me stumped. To begin, the whole equation is divisible by 3, so that gives us:

-x^2  +  12x  -  50

X Intercepts on that are a pain to figure out. I think that it can be expressed in turning point form as;

-x^2 + 12x +36 -36 -50
or,
-[x^2 + 12x + 36] - 36 - 50
therefore,
-(x+6)^2 - 36 - 50
simplified
-(x+6)^2 - 86

given that the number is -86 doesn't have a square root, it has to be expressed as a surd and put into the equation as -(x+6+SURD)(x+6-SURD) which will give the x intercepts, and from there turning point can be figured out.

From here, idk exactly what to do.. and I'm not sure that my method is correct.

Anybody know the X Intercepts and the turning point?

#2 frivolity

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Posted 07 October 2009 - 12:56 AM

The mistake you made is here:

-x^2 + 12x +36 -36 -50
or,
-[x^2 + 12x + 36] - 36 - 50

The negative sign cannot be taken out of the brackets without changing all the positives to negatives. It should become:

-[x^2 - 12x - 36] - 36 - 50, which has a quadratic within the parantheses that cannot be changed into a square.

The easiest method to solve this would be to divide everything by -3 thus yielding x^2 - 12x + 50 = 0

=> x^2 - 12x + 36 - 36 + 50 = 0
=> (x - 6)^2 = -14 => There are no x-intercepts.

Since -3(x - 6)^2 - 42 = 0, the turning point would be at x = 6. (Set the term in parantheses equal to zero).

Edit: To expand a little bit on having no x-intercepts, the graph is a "frowning" curve that is fully underneath the the x-axis and thus will not have any x-intercepts. There are possible imaginary roots for x though, but I don't think that the question requires that. If imaginary numbers are needed, then the solutions will be (root 14)i + 6 and -(root 14)i + 6.

Edited by frivolity, 07 October 2009 - 01:20 AM.





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