1. The equation
of a curve is

*y*= 10 –*x*^{2}+ 6*x*.
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 –

*x*^{2}+ 6*x*=*a*– (*x*+*b*)^{ 2}
10 –

*x*^{2}+ 6*x*=*a*– (*x*^{2}*+ 2bx +**b*^{2})^{ }; expand the squared term
10 –

*x*^{2}+ 6*x*=*a*–*x*^{2}*– 2bx –**b*^{2}; distribute the – sign
–

*x*^{2}+ 6*x*+ 10 = –*x*^{2}*– 2bx –**b*^{2}+ a ; order terms by their degree
And match up the coefficients:

–

*x*^{2}= –*x*^{2}*; 6**x*= – 2bx à b = -3; 10 = –*b*^{2}+ a à a =10 +*b*^{2}= 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: x

^{2}– 8x + 12 > 0. Your answer must include a sketch.
Let y = P(x) = 1x

^{2}– 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*x*^{2}+ 2*x*+*k*= 3*kx*– 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 = x

^{2}+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 = 1x

^{2}+ (2 – 3k)x + (k +1)
y = ax

^{2}+ 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 = b

^{2}– 4ac
D = (2 – 3k)

^{2}– 4(1)(k + 1)
D = 4 – 12k + 9k

^{2}– 4k – 4
D = 9k

^{2}– 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.