Unit+4

4.1 Partial Factoring
Finding the Vertex Using the y Intercept and Symmetry

The y intercept has a "matching" point on the other side of the axis of symmetry. We can use the y intercept and this matching point to find the coordinates of the vertex.

y = ax(x + b/a) + c or y =x(ax + b) + c. The symmetric point is (b/a, c)
 * Partial factoring** is a method for finding two points on opposite sides of the of symmetry of a parabola. We use this when the standard form of the relation cannot be fully factored. For the quadratic relation y = ax 2 + bx + c, partial factoring will result in

Example.

A ball is throw up in the air. It's height in metres, after t seconds is H = -4.9t 2 + 39.2t + 1.75 What does it reach it's maximum height? What is its maximum height?

Partial Factor H = -4.9t(t - 8) + 1.75, the symmetry points are (0, 1.75) and (8, 1.75)

The midpoint is t= 4, therefore the ball we reach its maximum height in 4 seconds

Plug in t=4 to find maximum height. H = -4.9t 2 + 39.2t + 1.75 H = -4.9(4) 2 + 39.2(4) + 1.75 H = 80.15

Therefore the maximum height is 80.15 m

HW: p. 377 # 5 (omit iii), 9, 10, 12ab, 15

15. The underside of a bridge has the shape of a parabolic arch. It has a maximum height of 30m and a width of 50m. can a sailboat with a a a mast 27m above the water pass under the bridge 8 metres from the symmetry of the arch?

If we centre the parabola on the y-axis, the zeros are -25 and 25, the vertex is at 0, 30.

y = a(x - 25)(x+ 25) substitute in 0, 30 and solve for a 30 = a(0 - 25)(0 + 25) a = -6/125 y = -6/125(x - 25)(x + 25)

What is the height at x = 8. y = -6/125(8 - 25)(8 + 25) y = 26.926

Therefore the sailboat can't pass under the bridge at that point.

4.2 The Vertex Form of a Quadratic Relation.
y = a(x - h) 2 + k where the vertex is the point (h, k)

Advantages of vertex form. We can now read off the vertex directly from the equation. With the vertex and the 'a' value we can now draw a sketch of the parabola directly from the equation. To make the sketch better we can expand to find the y-intercept ah 2 + k

We can also determine the number of zeros. If a and k are the same sign, there are no zeros, if the signs are opposite there are two zeros. If k is 0, then one zero.

We now have three forms of the quadratic relation, standard, factored and vertex.

To convert from vertex form to standard form, expand and collect like terms.

HW: p. 351 # 1, 2, 5ac, 7ac, 9, 10ac, 11.

4.3 Transformations of Quadratics Relations
y = x 2 is the base graph which we will transform in order to generate the graphs of all parabolas opening up or down.

Possible transformations of y = x 2 1. horizontal translation (slide) 2. vertical translation (slide) 3. vertical stretch/compression 4. vertical reflection (in the x axis) 5. combinations of # 1 - 4.

From y = a(x - h) 2 + k 1. The graph is moved h units to the right, (if h is negative, then hortizontal translation to the left) 2. The graph is moved k units up (if k is negative, then the vertical translation is down) 3. if the absolute value of a is greater than 1, then a vertical stretch (narrow), if absolute value less than 1, then vertical compressed (flatter). 4. If a is negative then there is a reflection on the x-axis

[|Transformation Website] HW: p. 363 # 1, 2, 9, 16.
 * Please study p360** in the text for visual representations and fuller explanations.

4.4 Converting to Vertex Form by Completing the Square
Steps:

1) Remove the common constant term from both x 2 and x.

2) Find the constant that must be added and subtracted to create a perfect square. (b/2) 2

3) Add and subtract this value after the x-term inside the brackets

4) Group the three terms form the perfect square. Move the subtracted value outside the brackets by multiplying it by the common constant factor

5) Factor the perfect square and collect like terms.

Example: 2x 2 - 10x + 7 =2[x 2 - 5x] + 7 =2[x 2 - 5x + 25/4 - 25/4] + 7 =2(x 2 - 5x + 25/4) - 25/2 + 7 =2 (x - 5/2) 2 +11/2 Vertex is (5/2, 11/2)

HW: p. 390 # 1adg, 2adg, 4adg, 6adg

4.5 The Quadratic Formula
[|intMath Quadratic Formula]

HW p403 #4, #6 every second question

4.6 The Discriminant
If b 2 - 4ac is greater than 0 there will be two real roots If b 2 - 4ac equal 0 there will be one root. (or some people say two equal roots) If b 2 - 4ac is less that 0 there are no real roots (or two imaginary roots)

p403 #5

Unit Test Wednesday May 12th
1. Converting to vertex form Method 1 - Factoring, finding the zeros, and finding the vertex from the midpoint Method 2 - Partial Factoring Method 3 - Completing the square

You need to know each method, in particular completing the square.

2. Transformations - There are up to four transformations. You should be a able to graph a parabola from an equation in vertex form.

3. Quadratic Formula and the Discriminant.

You also need the skills you learned in Unit 3.

Word Problems
1. You are given the zeros and a point. Make an equation in factored form and solve for 'a'. 2. You are given the vertex and a point. Make an equation in vertex form and solve for 'a'. 3. You are given an equation. Factor or use quadratic formula to find zeros. Partial Factor or complete the square to find vertex. 4. As the price goes up, the quantity goes down. Let x be the number of price increases. Make a revenue equation, this is already in factored form, so solve for the zeros. Find vertex by using midpoint. 5. Area of a matt/picture. Let x be the width of the border. Gives you an equation that needs to be expanded and simplified. 6. Given two numbers.... Let x be one number. Find the relation to the other number. Make an equation. 7. Pythagorean Theorem problem. Let x be one side, use Pythagorean theorem to make equation.

The first three types are really fundamental to the course.