Alan

This can be written as y = -x + 1      It has a slope of -1 and a y-intercept of 1

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What do you want to do with this?   It doesn't factor nicely, so perhaps you want to find the values of x that make the expression equal zero.  If so you could use the quadratic equation \(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\) or complete the square.

Set the expression equal to 0 and divide through by -12:  \(x^2 + \frac{x}{12}-\frac{7}{12}=0\)

Add \((\frac{1}{2\times12})^2 \rightarrow \frac{1}{576}\) to both sides and add 7/12 to both sides:

\(x^2+\frac{x}{12}+\frac{1}{288}=\frac{1}{576}+\frac{7}{12}\)   

The left hand side is now a perfect square and, since 7/12 = 376/576 we can write  \((x+\frac{1}{24})^2=\frac{337}{576}\)

Take the square root of both sides:   \(x=-\frac{1}{24}\pm\frac{\sqrt{337}}{24}\)

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This was answered here: http://web2.0calc.com/questions/show-step-please#r1 

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A general quadratic can be written as  y = ax² + bx  + c  where a, b and c are constants that can be found by plugging in the x and y values of the points specified:

4 = 4a + 2b + c          (1)

0 = 4a - 2b + c           (2)

0 = 16a + 4b + c       (3)

Subtract (2) from (1):     4 = 4b    so   b = 1

Subtract (2) from (3):     0 = 12a + 6b

 or   0 = 12a + 6

or  12a = -6                so  a = -1/2

 Put a and b back into (2):    0 = 4*(-1/2) -2*1 +c

or 0 = -4 + c   so  c = 4  

The quadratic is y = -x²/2 + x + 4     (you should check that this goes through the three specified points by substituting the x values in and checking if the right y values appear).

The first one exists:

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The second one doesn't exist as the numerator tends to a finite value and the denominator tends to 0.  Division by zero is undefined.

The standard deviation can be expressed as follows

In your case n = 12;   \(\mu=48/n\)  and the sum of the x² terms is 192.38.

I'll leave you to do the number crunching!

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As follows:

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Here's my version of the second one:

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Here's the first one:

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Divide by 2 first and take constant to the RHS:  \(x^2+3x=\frac{5}{2}\)

Now add square of half the coefficient of x to both sides:  \(x^2+3x+\frac{9}{4}=\frac{5}{2}+\frac{9}{4}\)

Rewrite as:  \((x+\frac{3}{2})^2=\frac{19}{4}\)

You should be able to take it from here.