Unit 2 Quadratics

Mrs. Valentine Math 3

2.1 Factoring and the Quadratic Formula•  Factoring ax2 + bx + c when a = ±1

•  Reverse FOIL method •  Find factors of c that add up to b. •  Using the factors, write the expression as the product of two

binomials. •  Example: Factor x2 + 9x + 20

•  Finding Common Factors •  Determine the greatest common factor •  Factor out the GCF from each term •  Use the distributive property •  Examples: Find the GCF and factor the expression. 6n2 + 9n 4x2 + 20x – 56

Factors of 20 Sum of Factors (9)

1 & 20 2 & 10 21 12

4 & 5 9

= (x + 4) (x + 5)

= 3n (2n + 3) = 4 (x2 + 5x – 14)

2.1 Factoring and the Quadratic Formula•  Factoring ax2 + bx + c when |a| ≠ 1

•  Reverse FOIL •  Multiply a and c •  Find factors of ac that add up to b •  Replace the bx term with the addition of the two factors (each

followed by x). •  Find the GCF of the first two terms and of the second two terms. •  Rewrite using the Distributive Property •  Examples: 4x2 + 7x + 3 2x2 – 7x + 6

ac = 4*3 = 12

= (2x2 – 4x) + ( –3x + 6) Factors of 12 Sum of Factors (7)

1 & 12 2 & 6 13 8 7

3 & 4

= (4x2 + 4x) + (3x + 3)

= 4x(x + 1) + 3(x + 1)

= (4x + 3) (x + 1)

ac = 2*6 = 12

= 2x(x – 2) – 3( x – 2)

= (2x – 3) ( x – 2)

2.1 Factoring and the Quadratic Formula•  Factoring a Perfect Square Trinomial

•  a2 + 2ab + b2 = (a + b)2 and a2 – 2ab + b2 = (a – b)2 •  Are the first and third terms perfect squares? •  Does the middle term equal twice the product of their square roots? •  What is the sign of b? •  Apply the rule for factoring. •  Examples: 4z2 – 20z + 25 4x2 + 12x + 9

•  Factoring a Difference of Squares •  a2 – b2 = (a + b) (a – b) •  Take the square root of each term. •  Add them for one factor and subtract them for the second. •  Examples: 25x2 – 49 p2 – 81

= (2z – 5)2 = (2x + 3)2

= (5x – 7)(5x + 7) = (p + 9)(p – 9)

2.1 Factoring and the Quadratic Formula•  Using the Quadratic Formula

•  Write the equation in standard form •  Find the values of a, b, and c. •  Plug in to the quadratic formula •  Simplify •  Examples: x2 + 6x + 9 = 0 m2 + 8m + 12 = 0

•  Using the Discriminant •  b2 – 4ac, can be positive, zero, or negative

•  Positive: 2 real solutions •  Zero: 1 real solution •  Negative: no real solutions, 2 imaginary solutions

•  Examples: Determine the number of real solutions. –x2 + 2x – 9 = 0 k2 – 6k + 9 = 0

for ax2 + bx +c

x = –3 m = – 6, –2

No real solutions

1 real solution

2.2 Completing the Square•  Solving Using Square Roots

•  Set y = 0 and isolate x2 •  Take the square root of both sides •  You should get both a positive and negative answer. •  Examples: x2 – 36 = y n = b2 – 13

•  Using Zero Product Property

•  Factor the quadratic function •  Set each factor equal to zero •  Solve each for the variable •  Examples: w2 – 19w – 20 = 0 2r2 – r – 6 = 0

x2 – 36 = 0

x2 = 36 x = –6, 6

0 = b2 – 13 13 = b2

–√(13), √(13) = b

w = –1, 20 r = –3/2, 2

2.2 Completing the Square•  Determining Dimensions

•  While designing a house, an architect used windows like the one shown here. What are the dimensions of the window if it has 2766 square inches of glass?

A = l*w

= (2x)*(x) A = ½ πr2

= ½ π(x/2)2 = ½ π(x2/4) = 2x2

= π(x2/8)

2766 = 2x2 + π(x2/8) 2766 = x2 (2+ (π/8))

2766 / (2+ (π/8)) = x2

√(2766 / (2+ (π/8))) = x

±34 ≈ x

The width of the window is 34in while

the height is 68in. The semicircular top has a

radius of 17in.

2.2 Completing the Square•  Solving Perfect Square Trinomials

•  Factor the perfect square trinomial. •  Find the square roots. •  Rewrite as two equations. •  Solve for the variable. •  Examples:

x2 + 4x + 4 = 25 9p2 + 24p + 16 = 36

(x + 2)2 = 25

x + 2 = ±√(25)

x + 2 = ±5 x + 2 = 5 x+2 = –5

x = 3 x = –7

(3p + 4)2 = 36

3p + 4 = ±√(36)

3p + 4 = ±6 3p + 4 = 6 3p + 4 = –6

3p = 2 3p = –10 p = 2/3 p = –10/3

2.2 Completing the Square•  Completing the Square

•  When the x2 + bx is not part of a perfect square trinomial, we can use completing the square to find solutions.

•  Examples: 3u2 – 12u + 6 = 0

•  Divide all terms by a •  Move c to the other side of the

equation. •  Find (b/2)2. •  Add it to both sides of the equation. •  Factor and solve the perfect square trinomial.

u2 – 4u + 2 = 0

u2 – 4u = –2

(b/2)2 = (4/2)2 = 22

u2 – 4u + 22 = – 2 + 22

(u – 2)2 = 2

u – 2 = ±√(2)

u – 2 = √(2) u – 2 = –√(2)

u = 2 + √(2) u = 2 – √(2)

2.2 Completing the Square•  Writing in Vertex Form

•  Both add and subtract (b/2)2 •  Factor the perfect square trinomial •  Simplify •  Examples:

y = x2 + 4x – 6 2x2 – 6x – 1 = y

(b/2)2 = (4/2)2 = 22

y = x2 + 4x + 22 – 6 – 22

y = (x2 + 4x + 22) – 6 – 4

y = (x + 2)2 – 10

(b/2)2 = (-3/2)2 = (3/2)2

x2 – 3x – ½ = (½)y

x2 – 3x + (3/2)2 – ½ – (3/2)2 = (½) y

(x2 – 3x + (3/2)2) – 2/4 – 9/4= (½) y

(x – (3/2))2 – 11/4= (½) y

2(x – (3/2))2 – 11/2 = y

2.3 Parabolas (Focus and Directrix)•  Parabolas with Equation y = ax2 (vertical parabolas)

•  A parabola is the set of all points in a plane that are the same distance from a fixed line and a fixed point not on the line.

•  Fixed point = focus (0,p) •  Fixed line = directrix (y = –p) •  Distance between vertex and focus = focal length

•  Equation of the vertical parabola: •  can be used to find the focus and the

directrix •  Examples: Find the equation of the parabola with vertex (0,0)

and F(0, -1.5).

•  What are the focus and directrix of ?

The focus is (0,–3). The directrix is y = 3.

y = –1/6x2

2.3 Parabolas (Focus and Directrix)•  Using Parabolas to Solve Problems

•  Geometry of parabolas gives real-world meaning to focus. A light shone from the focus point reflects along parallel lines. This is how a flashlight works.

•  The parabolic solar reflector pictured has a depth of 2 feet at the center. How far from the vertex is the focus? What is the focal length?

Start by graphing the parabola.

Using the general form of the equation, substitute one of the points.

The focus is (0,2) The focal length is 2ft.

2.3 Parabolas (Focus and Directrix)•  Analyzing a Parabola

• 

•  Find the completed square form of the equation, then find p, h, and k •  Use these values to determine vertex, focus, and directrix. •  Examples: y = x2 – 4x + 8 y = 2x2 + 4x – 2

•  Writing an Equation of a Parabola •  Given a focus and the vertex, you can write the equation of a

parabola. •  Examples: vertex: (7, 2) focus (7, –2)

vertex (-5, 4) focus (-5, 0)

Vertical Parabola Vertex (0,0) Vertex (h,k) Equation y = (1/4p)x2 y = (1/4p)(x – h)2 + k Focus (0, p) (h, k + p) Directrix y = –p y = k – p

y = (–1/16)(x – 7)2 + 2

y = (–1/16)(x + 5)2 + 4

Vertex (2,4); focus (2, 4.25); directrix y = 3.75

Vertex (–1, –4); focus (–1, –3.875); directrix y = –4.125

2.3 Parabolas (Focus and Directrix)•  Parabolas with Equation x = ay2 (horizontal parabolas)

•  this is the same as it was for vertical parabolas

•  Focus: F(p,0) •  Directrix: x = –p

•  Examples: Write the equation of a parabola with vertex at the origin and directrix x = 1.25.

•  What are the vertex, focus, and directrix of the parabola with equation x = 0.75y2?

•  Horizontal Parabola Vertex (0,0) Vertex (h,k) Equation x = (1/4p)y2 x = (1/4p)(y – k)2 + h Focus (p, 0) (h + p, k) Directrix x = –p x = h – p

x = –1/5y2

Vertex (0,0); focus (1/3, 0); directrix x = –1/3

2.3 Parabolas (Focus and Directrix)Horizontal Parabola Examples

•  What are the focus and directrix of x = (1/16)y2 ?

•  Find the vertex, focus, and directrix for –8x = y2 – 8y +16.

•  Find the equation of a horizontal parabola with vertex (3,2) and focus (4,2).

vertex (0,0); focus (4,0); directrix: x = –4

vertex (0, 4); focus (–2,4); directrix x = 2

x = ¼ (y – 2)2 + 3

2.4 Circles in the Coordinate Plane•  Standard Form of the Equation of a Circle

•  center (h,k) and radius = r

•  Examples: Give the equation of the circle with the provided center and radius. center (–4, 3) radius = 4

center (–2, –6) radius = 7

•  Writing Equations Using Translations •  Write the equation given •  Replace h and k with the horizontal and vertical translations •  Simplify. •  Example: x2 + y2 = 9 translated 4 units left and 3 units up.

2.4 Circles in the Coordinate Plane•  Using Graphs to Write Equations

•  Determine the radius, r •  Find the center (h, k) •  Substitute r, h, and k into the equation for a circle. •  Examples:

•  Finding the Center and Radius •  Rewrite the equation in standard form •  Find h, k, and r •  Example: (x – 16)2 + (y + 9)2 = 144

r = 3 h = -3 k = 4

r = 4 h = 2 k = -6

(x – 16)2 + (y – (– 9))2 = 122

r = 12 h = 16 k = -9

Center: (16, -9) Radius: 12

2.4 Circles in the Coordinate Plane•  Example: x2 + y2 – 6x + 14y = 6

•  Graphing Circles •  You can use the center and the radius to graph a circle. •  Steps:

•  Write the equation in standard form. •  Identify h, k, and r to get center and radius. •  Plot the center. •  Find (and plot) 4 points distance r away from the center •  Draw the circle using the points you have plotted

(x2 – 6x) + (y2 + 14y) = 6

(x2 – 6x + 9) + (y2 + 14y + 49) = 6 + 9 + 49

(x – 3)2 + (y + 7)2 = 64

(x – 3)2 + (y – (–7))2 = 82 r = 8 h = 3 k = -7

Center: (3, -7) Radius: 8

(6/2)2 = 32 = 9 (14/2)2 = 72 = 49

2.4 Circles in the Coordinate Plane•  Examples:

(x + 1)2 + (y – 3)2 = 25 (x – 7)2 + (y – 1)2 = 100

C: (– 1, 3) r = 5 C: (7, 1) r = 10