## Tuesday, October 30, 2012

1.  The equation of a curve is y = 10 – x2 + 6x.
a) Express y in the form a – (x + b) 2 , where a and b are integers.
The equation for y in the expanded form = The equation for y in the given form
10 – x2 + 6x = a – (x + b) 2
10 – x2 + 6xa – (x2 + 2bx +  b2) ; expand the squared term
10 – x2 + 6xa –  x2 –  2bx –  b2; distribute the  –  sign
x2 + 6x + 10 =  –  x2 –  2bx –  b2 + a ; order terms by their degree
And match up the coefficients:
x2 =  –  x2 ;  6x =  –  2bx  à b = -3;  10 =  –  b2 + a à a =10 + b2 = 19
b) Hence, find the coordinates of the vertex of the curve.
The vertex form is  Y = A(X – H)2 + K
The vertex form is also Y =  K + A(X – H) 2 by the Commutative Law of Addition (e.g., 2 + 3 = 3 + 2)
Compare with                y = 19 – (x + (-3)) 2
Obviously, K = 19,    and A = -1.
Also – H = +(-3)
Therefore H = +3
Hence, the vertex is (x, y) = (H, K) = (+3, +19)

2. Solve the quadratic inequality:   x2 – 8x + 12 > 0.     Your answer must include a sketch.
Let y = P(x) = 1x2 – 8x + 12, with a = 1, i.e. a positive number. We have no requirement for b and c.
To find the roots change the form of the polynomial from expanded form to factored form.
y = (x – 6)(x – 2) in factored form
Set y = 0 to find the roots.
x – 6 = 0 à x = 6
x – 2 = 0 à x = 2.   The roots are “critical numbers” on the x-axis.  Between the roots the parabola is under the x-axis. To the left and right of the roots the parabola lies above the x-axis. Check the original problem.  We need the 2 separate pieces of the parabola above the x-axis.

Using our knowledge of set theory, we can write the solution as {x: x < -2 U x > 6, x is a Real number}.  The solution set is composed of 2 disjoint pieces.

3. Determine the set of values of k for which the equation x2 + 2x + k = 3kx – 1 has no real roots.
As given, this equation is of the form, quadratic function = linear function.
A rough sketch shows that there can be 2, 1 or zero solutions. (The sketch of a quadratic function is a parabola. The sketch of a linear function is a line. Amazing!)

To solve, we bring all the terms to the left.
We need (quadratic) – (line)  = 0  has no real roots.
New quadratic = 0 has no real roots
Do not make the mistake of believing that in the equation y = x2 +2x 3kx + k + 1that ‘x’ and ‘k’ are both variables!!!  Yes, ‘x’ is a variable, because we can draw an x-axis.  However, k is merely a parameter.
y = 1x2 + (2 – 3k)x + (k +1)
y = ax2 +            bx + c.
By direct comparison:  a = 1, b = (2 – 3k) and c = (k + 1)
The phrase “no real roots” should tell you to use the discriminant.
Discriminant = b2 – 4ac
D = (2 – 3k) 2 – 4(1)(k + 1)
D = 4 – 12k + 9k2 – 4k – 4
D = 9k2 – 16k in expanded form or D = k(9k – 16) in factored form.
That is, D is a polynomial function of k.  In fact this will also be a parabolic function.
Now, we can give D and k their own axes.
Using set notation the solution is { 0 < k < 16/9}.   Unlike the previous question, all the points in this set form one continuous piece.