Quadratic Functions Test

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 + 6x =  a – (x² + 2bx +  b²) ; expand the squared term

10 – x² + 6x =  a –  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.