Quadratic Functions. Expanding to Standard Form A quadratic function is a function that can be written in the standard form below and where quadratic.

Quadratic Functions

Expanding to Standard Form

A quadratic function is a function that can be written in the standard form below and where quadratic term linear term constant term If a = 0, then the function has no quadratic term & it is not a quadratic function.

F.O.I.L. First,Outer,Inner,Last

 Simplify a.) First,Outer,Inner,Last Combine Like Terms!!

b.) First,Outer,Inner,Last Combine Like Terms!!

 Expand to standard form & determine whether each function is linear or quadratic. a.) b.)

c.) d.) e.)

 Worksheet: Expanding to Standard Form

Quadratic Graphs

 The graph of a quadratic function is a parabola.  The axis of symmetry is the line that divides a parabola into two parts that are mirror images. axis of symmetry

 The vertex of a parabola is the point at which the parabola intersects the axis of symmetry.  The y-value of the vertex represents the maximum or minimum value of the function. vertex

 If the x 2 term is positive, the parabola is concave up & the vertex is the minimum point  If the x 2 term is negative, the parabola is concave down & the vertex is the maximum point

 Identify the vertex, tell whether it’s a minimum or a maximum, identify the axis of symmetry, & describe the graph’s concavity. a.) Vertex: Minimum b.) Vertex: Maximum Axis of Symmetry: x=1 Concave Up Axis of Symmetry: x=-4 Concave Down

xy=1/2x 2 y  Make a table of values & graph the function  Draw in your axis of symmetry  Reflect the points across the axis of symmetry & draw line a.) xy=1/2x 2 y 0 1 2 x y 0y=1/2(0) 2 1 2 xy=1/2x 2 y 0y=1/2(0) 2 0 1 2 xy=1/2x 2 y 0y=1/2(0) 2 0 1y=1/2(1) 2 2 xy=1/2x 2 y 0y=1/2(0) 2 0 1y=1/2(1) 2 1/2 2 xy=1/2x 2 y 0y=1/2(0) 2 0 1y=1/2(1) 2 1/2 2y=1/2(2) 2 xy=1/2x 2 y 0y=1/2(0) 2 0 1y=1/2(1) 2 1/2 2y=1/2(2) 2 2

xy=2x 2 y b.) xy=2x 2 y 0 1 2 x y 0y=2(0) 2 1 2 xy=2x 2 y 0y=2(0) 2 0 1 2 xy=2x 2 y 0y=2(0) 2 0 1y=2(1) 2 2 xy=2x 2 y 0y=2(0) 2 0 1y=2(1) 2 2 2 xy=2x 2 y 0y=2(0) 2 0 1y=2(1) 2 2 2y=2(2) 2 xy=2x 2 y 0y=2(0) 2 0 1y=2(1) 2 2 2y=2(2) 2 8  Make a table of values & graph the function  Draw in your axis of symmetry  Reflect the points across the axis of symmetry & draw line

+ up - down narrow/wide translates (shifts) graph up/down

xy=x 2 -4y  Graph the function xy=x 2 -4y 0 1 2 x y 0y=(0) 2 -4 1 2 xy=x 2 -4y 0y=(0) 2 -4-4 1 2 xy=x 2 -4y 0y=(0) 2 -4-4 1y=(1) 2 -4 2 xy=x 2 -4y 0y=(0) 2 -4-4 1y=(1) 2 -4-3 2 xy=x 2 -4y 0y=(0) 2 -4-4 1y=(1) 2 -4-3 2y=(2) 2 -4 xy=x 2 -4y 0y=(0) 2 -4-4 1y=(1) 2 -4-3 2y=(2) 2 -40

 Worksheet: Quadratic Graphs

Graph y = ax 2 +bx+c

Axis of Symmetry This will also give you the x-coordinate of the vertex

 Graph the function a.) 1.) Find the axis of symmetry/vertex Plug in x-coordinate to get y-coordinate Vertex =(1,8)

2.) Find the y-intercept Plug in 0 for x to find y-intercept y-intercept =(0,5)

3.) Find a 3 rd point that is on the same side of the axis of symmetry as the y-intercept Could plug in for x:-1, -2, -3, etc Let x = -1 3 rd Point:(-1,-4)

4.) Reflect points across axis of symmetry & draw graph

 Graph the function b.) 1.) Find the axis of symmetry/vertex Plug in x-coordinate to get y-coordinate Vertex =(3,0)

2.) Find the y-intercept Plug in 0 for x to find y-intercept y-intercept =(0,9)

3.) Find a 3 rd point that is on the same side of the axis of symmetry as the y-intercept Could plug in for x:2, 1, -1, etc Let x = 1 3 rd Point:(1, 4)

4.) Reflect points across axis of symmetry & draw graph

 Graph c.) 1. Axis of Sym.: Vertex = (-1, -6) 2. y-intercept: y-intercept = (0, -5) 3. 3 rd Point: Let x = 1 3 rd Point = (1, -2)

 Graph d.) 1. Axis of Symm: Vertex = (2, 0) 2. y-intercept: y-intercept = (0, -4) 3. 3 rd Point: Let x = 1 3 rd Point = (1, -1)

Solving Quadratics Equations

 Recall: Standard Form of Quadratic Equation  To solve a quadratic equation, we want to know when y = 0; therefore, we set the equation equal to 0  Imagine the graph of a quadratic equation. What happens when y = 0? OR x-intercepts! The solutions to a quadratic equation are the x-intercepts!!

xy  Solve by graphing a.) Treat it as if it’s y=x 2 -4 xY 0 1 2 xy 0-4 1 2 xy 0 1-3 2 xy 0-4 1-3 20 y = 0 at the x-intercepts, which are: -2 & 2 Therefore, the solutions are: -2 & 2

xy  Solve by graphing b.) xY 0 1 2 xy 0-9 1 2 xy 0 1-8 2 xy 0-9 1-8 2-5 The graph doesn’t cross the x-axis. But, will it eventually? The solutions are:-3 & 3 Add more #’s to your table! xy 0-9 1-8 2-5 3 xy 0-9 1-8 2-5 30

xy  Solve by graphing c.) xY 0 1 2 xy 04 1 2 xy 04 15 2 xy 04 15 28 Therefore, the solution is: No solution The graph doesn’t cross the x-axis. But, will it eventually?

Some equations can be solved by simply solving for the variable

 Solve using square roots a.) Solve for x! To un-do squaring of x, square root! (must get x 2 alone first) Be sure to include both solutions!

b.)c.)

 Expand to standard form An equation in this form expands to a quadratic equation. Therefore, this is a quadratic equation in factored form

For every real number, a & b, a = 0 or b = 0if ab = 0, then Ex:If (x+3)(x+2) = 0, then x+3 = 0 or x+2 = 0

 Solve a.) OR OR b.) OR OR

 Worksheet: Solving Quadratic Equations

Factoring x 2 +bx+c

Since 3 x 5 = 15, 3 & 5 are factors of 15 Since, & are factors of *What do you notice about 3 & 5?

 Factor a.) Find factors of +12 that add up to +7 12 1 & 12 2 & 6 3 & 4 1+12 = 13 2+6 = 8 3+4 = 7 (x+ )(x+ )34

b.) Find factors of +30 that add up to +13 30 1 & 301+30 = 31 2 & 152+15 = 17 3 & 103+10 = 13 (a+ )(a+ ) 3 10

c.) Find factors of +42 that add up to -17 42 -2 & -21-2+(-21) = -23 -3 & -14-3+(-14 )= -17 (d- )(d- )314 Both factors will have to be negative!

d.) Find factors of +18 that add up to -11 18 -1 & -18-1+(-18) = -19 -2 & -9-2+(-9) = -11 (x- )(x- ) 29 Both factors will have to be negative!

e.) (m+ )(m- ) 39 -27 -1 & 27 -1+27 = 26 -3 & 9-3+9 = 6 Find factors of -27 that add up to +6 One will have to be positive and one has to be negative. The larger factor will need to be positive!

f.) -18 1 & -18 1+(-18) = -17 2 & -9 Find factors of -18 that add up to -3 2+(-9) = -7 3 & -63+(-6) = -3 (p+ )(p- )36 One will have to be positive and one has to be negative. The larger factor will need to be negative!

g.) Find factors of -20 that subtract to get +8 -20 -1 & 20 -1+20 = 19 -2 & 10 -2+10 = 8 (m+ )(m- )102 One has to be positive and one has to be negative. The larger factor will need to be positive!

h.) Find factors of -56 that add up to -1 -56 1 & -56 1+(-56) = -55 2 & -282+(-28) = -26 4 & -14 4+(-14) = -10 7 & -87+(-8) = -1 (y+ )(y- )87 One has to be positive and one has to be negative. The larger factor will need to be negative!

 Worksheet: Factoring Day 1 x 2 +bx+c

Factoring ax 2 +bx+c

 Factor a.) 1.) Multiply the 1 st & last terms, find factors of that # that combine to get middle term Factors of 42: 1 & 421 + 42 = 43 2 & 212 + 21 = 23 2.) Split the middle term 3.) Factor out the GCF from the 1 st two terms, then the last two terms

b.) Factors of -56: 1 & -561 + (-56) = -55 2 & -282 + (-28) = -26 1.) Multiply the 1 st & last terms, find factors of that # that combine to get middle term 2.) Split the middle term 3.) Factor out the GCF from the 1 st two terms, then the last two terms

C ONTINUED … c.) Factors of 4: -2 & -2-2 + (-2) = -4 -1 & -4-1+ (-4) = -5 1.) Multiply the 1 st & last terms, find factors of that # that combine to get middle term 2.) Split the middle term 3.) Factor out the GCF from the 1 st two terms, then the last two terms

C ONTINUED … d.) Factors of -6: -1 & 6-1+ 6 = 5 -2 & 3-2+ 3 = 1 1.) Multiply the 1 st & last terms, find factors of that # that combine to get middle term 2.) Split the middle term 3.) Factor out the GCF from the 1 st two terms, then the last two terms

C ONTINUED … e.) 2.) Multiply the 1 st & last terms, find factors of that # that combine to get middle term Factors of 28: 1 & 281+ 28 = 29 2 & 142+ 14 = 16 3.) Split the middle term 4.) Factor out the GCF from the 1 st two terms, then the last two terms 1.) First, factor out the GCF Now, just factor what’s left in brackets & just carry down the 5

C ONTINUED … f.) 2.) Multiply the 1 st & last terms, find factors of that # that combine to get middle term Factors of 5: -1& -5-1+ -5 = -6 3.) Split the middle term 4.) Factor out the GCF from the 1 st two terms, then the last two terms 1.) First, factor out the GCF Now, just factor what’s left in brackets & just carry down the 2

 Worksheet: Factoring Day 2 ax 2 +bx+c

Factoring a Difference of Two Squares

 Factor a.) b.) c.) d.) e.) f.)

 Worksheet: Factoring Day 3 All types

Solving Quadratics by Factoring

 Solve using factoring a.) (x )(x ) = 0 OR OR b.) OR

a.) Must get into standard form first! OR

 Worksheet: Solving Quadratics by Factoring

Simplifying Square Roots

 Square Roots 2 We’re taking the square root, therefore, the index is 2. We just don’t write it.

Multiplication Property of Square Roots Ex:

 Simplify a.) Step 1: List factors of 54, find pair with largest perfect square 54 2 & 27 3 & 18 6 & 9 Step 2: Break up the square root into its two factors (write perfect square first) Step 3: Simplify b.) 50 2 & 25 5 & 10

C ONTINUED … d.) 18 2 & 9 3 & 6 c.) 500 2 & 250 5 & 100 25 & 20

 Worksheet: Simplifying Radicals

Using the Quadratic Formula

If ax 2 +bx+c = 0, and a 0, then *Equation always has to be in standard form first

 Solve a.) Already in standard form a = -3b = 5c = -2 OR

 Solve b.) Put into standard form a = 1b = -5c = 6 OR

 Solve c.) Already in standard form a = 2b = 6c = 1

 Solve d.) Put into standard form a = 2b = 4c = -7

 Worksheet: Using the Quadratic Formula

Completing the Square

You can solve equations in which one side is a perfect square trinomial by taking the square root of each side.

 Solve The left side is a perfect square trinomial, so factor! Square root each side OR

If one side of an equation is not a perfect square trinomial, you can convert it into one by rewriting the constant term. This is called completing the square. Use the following relationship to find the term that will complete the square.

 Find the missing value to complete the square a.)

 Solve by completing the square a.) 1.) Find 2.) Re-write so all terms containing x are on one side 3.) Complete the square by adding 36 to each side 4.) Factor 5.) Square root each side 6.) Solve for x

b.) 1.) Find 2.) Re-write so all terms containing x are on one side 3.) Complete the square by adding 4 to each side 4.) Factor 5.) Square root each side 6.) Solve for x

c.) Since there is a value for a that is greater than 1, the order of steps changes! 1.) Re-write so all terms containing x are on one side 2.) Divide both sides by the coefficient of x 2 term 3.) Find 4.) Complete the square by adding 1 to each side 5.) Factor 7.) Solve for x 6.) Square root both sides

 Worksheet: Completing the Square

Translating Parabolas: Part I

Just as we graphed absolute value functions as translations of their parent function, we can graph a quadratic function as a translation of the parent function.

To translate the graph of a quadratic function, we’ll use the vertex form of a quadratic function. - Right + Left + Up - Down The vertex is (h, k) & the axis of symmetry is the line x=h

 Graph Step 1: Graph the vertex Step 2: Draw the axis of symmetry Step 3: Find & graph the y-intercept Step 4: Find 3 rd pointLet x = -2 Step 5: Reflect points & draw graph

 Graph Step 1: Graph the vertex Step 2: Draw the axis of symmetry Step 3: Find & graph the y-intercept Step 4: Find 3 rd pointLet x = 1 Step 5: Reflect points & draw graph

 Write the equation of the parabola Vertex: (3, 4)Other point: (5, -4) h, k x, y Use vertex form Substitute for h, k, x, & y Solve for a Write equation in vertex form, using values for a, h, & k

 Write the equation of the parabola Vertex: (-1, 0)Other point: (-2, 2) h, k x, y Use vertex form Substitute for h, k, x, & y Solve for a Write equation in vertex form, using values for a, h, & k

 Textbook p. 251 #2-20 even (printout) Use Graph paper!!

Translating Parabolas: Part II

 Write in vertex form a.) 1.) Find the x-coordinate of the vertex 2.) Plug that value in to get the y-coordinate of the vertex Vertex is (2, 17) 3.) Plug values for a (found in original equation), h, & k into vertex form h, k

b.) 1.) Find the x-coordinate of the vertex 2.) Plug that value in to get the y-coordinate of the vertex Vertex is (-2, -7) 3.) Plug values for a (found in original equation), h, & k into vertex form h, k

 Identify the vertex & the y-intercept of the graph of the function a.) hk Vertex is (1, -1) y-intercept: plug in 0 for x y-intercept is (0, 1)

b.) hk Vertex is (-2, 4) y-intercept: plug in 0 for x y-intercept is (0, -8)

 Textbook p. 251-252 (printout)  #1, 3, 17, 19, 21-24, 27-30 (Use Graph paper for #1 & #3)

Real-World Application

 The number of items a company sells frequently is a function of the item’s price.  The revenue from sales of the item is the product of the price and the number sold.

 The number of unicycles a company sells can be modeled by the function,, where p = price  What price will maximize the company’s revenue from unicycles?  What is the maximum revenue? Revenue =PricexNumber sold R =px(-2.5p+500)

 How can we find the maximum value of the function?  Since a<0, the graph opens down, and the vertex represents a maximum value.  Instead of (x, y), we’re using (p, r), so we need to find p at the vertex.  Now, find the value of r at the vertex The price of $100 will maximize revenue at $25,000.

 The number of widgets the Woodget Company sells can be modeled by, where p = price.  What price will maximize revenue?  What is the maximum revenue? Revenue =Price x Number sold R =px(-5p+100) The price of $10 will maximize revenue at $500.

 Worksheet: Quadratics Real World Applications  Day 1

More Real-World Application

 Smoke jumpers are in free fall from the time they jump out of a plane until they open their parachutes.  The function models a jumper’s height in y feet at t seconds for a jump from 1600 ft.  How long is a jumper in free fall if the parachute opens at 1,000 ft? Plug in 1,000 for y & solve for t The jumper is in free fall for about 6.1 seconds

The figure shows a pattern for an open-top box. The total area of the sheet of material used to manufacture the box is 288 in 2. The length is 2 in longer than the width. The height of the box is 3 in. Therefore, 3in x 3in squares are cut from each corner. Find the dimensions of the box. Length x Width = Area 3 3 Define:x = width of a side of the box x Width of Material:3 + x + 3 = x+2 x + 6 Length of Material:3 + x + 2 + 3 =x + 8

Must equal 0 in order to factor (x )(x ) = 0 OR OR Not reasonable b/c negative Dimensions of Box: Width =x= 10 in. Length =x + 2 = 12 in. Height = 3 in

 The area of a square is  Find the length of a side. Is a perfect square trinomial Factor! The side of the square has a length of (3g + 2) cm

More Real-World Application

 The volume (lwh) of a rectangular prism is  Factor to find possible expressions for the length, width, & height of the prism. Factor out GCF Factor the quadratic trinomial 15 x 20 = 300 Factors of 300 that combine to get 56? The possible dimensions of the prism are 4x, (10x+3) & (2x+5) in

 Write in vertex form factor out -1 from 1 st two terms (in order to make x 2 positive) complete the square add & subtract 9 on the right side factor the perfect square trinomial Simplify

 Write in vertex form complete the square add & subtract 25 on the right side factor the perfect square trinomial Simplify

 The profit P from handmade sweaters depends on the price s at which each sweater is sold.  The function models the monthly profit from sweaters for one custom tailor.  Write the function in vertex form. Use the vertex form to find the price that yields the maximum monthly profit and the amount of the maximum profit. Factor -1 from the 1 st two terms Complete the square Add & subtract 3600 on the right side Factor the perfect Square trinomial Simplify in vertex form The vertex is (60, 1600), which means a price of $60 per sweater gives a maximum monthly profit of $1600.

Quadratic Functions. Expanding to Standard Form A quadratic function is a function that can be written in the standard form below and where quadratic.

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Quadratic Functions. Expanding to Standard Form A quadratic function is a function that can be written in the standard form below and where quadratic.

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Presentation on theme: "Quadratic Functions. Expanding to Standard Form A quadratic function is a function that can be written in the standard form below and where quadratic."— Presentation transcript:

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