Quadratic Relationships

Table of contents

Introduction

  • Definition
  • Vocabulary


Topic 1
  • First and second difference
  • Graphing vertex form and transformation / translations
  • Finding an equation given a point and vertex
  • Factored form graphing
  • Word problem


Topic 2


  • Multiplying binomials
  • Special products
  • Common factoring
  • Simple grouping
  • Factor Expressions of the form x2+bx+c
  • Factor Expressions of the form ax2+bx+c
  • Perfect square trinomial
  • Difference of square
  • Word problem


Topic 3


  • Standard form to vertex form
  • Solve quadratic equations
  • Graph quadratics using the x-intercepts
  • The quadratic formula
  • Word problem

Introduction

Quadratic relation corresponds to a quadratic formula which is y= ax2+bx+c. this formula is often used to calculate the height of falling rocks, kicked balls or an arched bridge.


Vocabulary

Parabola: is the curve formed from all the point of (x,y).

Vertex: the minimum or maximum point in the graph. It is the point where the graph changes directions.

Minimum/ maximum value: the lowest or highest point of the parabola.

Axis of symmetry: a line that goes down the middle of the parabola through the vertex.

Y-intercept: The point where the parabola crosses the y-axis.

X-intercept: The point where the parabola touches the x-axis.

Zeros: The point where the parabola crosses the x-axis ( Also called the roots).

Topic 1

First and second difference

To know if the graph will have a quadratic relation or linear relationship you find the first difference of the Y vales. To find the first or second difference you subtract the bottom number from the top.If the first difference is the same than it will be a linear relation. If the first difference is not the same, then find the second difference and if that is the same than the relationship will be quadratic.


Graphing vertex form

The equation of a parabola can be expressed as a standard form or vertex form. The h represents x and the k represents y. With that you have the vertex.

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Standard Form

The equation for standard form is y= ax2 + bx + c

If the "a" int he formula is positive than the parabola will open up. If the "a" is negative than the parabola will open downwards.

Vertex Form and transformation / translations

The equation for vertex form in y= a(x-h)2 + k

In this equation the "h" represents "x" and "k" represents "y".

If the "a" value is greater than 1 than the parabola will become narrow.

If the value of "a" is lower than 1 than the parabola will stretch.

If K>0, than the parabola will be translated k units up.

If K<0 than the parabola will be translated K units down.

If H<0 than the parabola will translate H units to the left.

If H>0 that the parabola will translate H units to the right.

Finding an equation given a point and vertex

To find an equation from a given point and vertex, you will use the vertex formula. First you substitute the X value of the point into the H. For example if your points are (-2, 8) and your x-intercept is (2,0). You use the first coordinate to substitute H and K. You use the second coordinate to find A.

Your equation would start as y= a(x+2)2 + k. You have to leave the negative sign the same unless the X value is negative, then it will turn positive.

To find your K, it will be the Y value.

So in this example the equation will become y= a(x+2)2 + 8. Your parabola will go to the left 2 units and up 8 units. Now to find A you will have to use the second coordinated which are (2,0). The 2 will take place of the X and the 0 will take place of the Y.

0=a(2+2)2 + 8. To find a you will have to isolate it.

0= a (4)2+8

0= a(16) + 8

-8 - 8

-8= 16a

16 16

-1/2= a

so the value of a is -1/2. now the full equation will be y= -1/2(x+2)2 + 8.

Factored form graphing


the equation you have to use to graph is standard from which is y=ax2+bx+c.

The equation could be y=x2-8x+12.

Since the value of a is positive the parabola will open upwards.

To Factor the trinomial, identify 2 numbers whose sum is –8 and the product is 12. The numbers are –2 and –6.

So the factored form would be (x-2)(x-6)=0

Now to find the zeroes you take each bracket and isolate the x.

(x-2)=0

for this you would add 2 on each side so it would look like

x=2. so the first zero is 2.

do the same to the second bracket and you will get 6. so the 2 zeros are 2 and 6.

To find the axis of symmetry you and the 2 zeros and divide them by 2.

2+6= 8

8/2=4. so the axis of symmetry is x=4. to find the vertex you have to sub in the value of x into the equation and you will get your y value.

y=x2-8x+12

y=(4)2-8(4)+12

Y=16-32+12

y=-4

so now you have the x-intercept as 2 and 6, you have the vertex as (4,-4) and you have the y-intercept which is 12. now you can plot the points and make a parabola.

Therefore, the x-intercepts of the function are 6 and 2.

The x-coordinate of the vertex is the midpoint of the x -intercepts. So, here the x-coordinate of the vertex will be .

Substitute x = 4 in the equation to find the y-coordinate of the vertex.

Word problem
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Topic 2

Multiplying binomials

To multiply binomials you take the first term in the first bracket and multiplied it with the first term in the second bracket and the second term in the second bracket. then you multiply the second term in the first bracket with the first and second term in the second bracket.

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the answer to this will be

x2+3x+2x+6

then you add the terms that are the same

x2+5x+6

this is how you multiply binomials.


Special Products

Special products are factored the same as multiplying binomials, the only difference is that there is only one equation and that equation is powered by 2.

An example is (x+4)2

It is the same concept as multiplying binomials but you just multiply the same equation with itself.

(x+4)(x+4)

x2+4x+4x+16

x2+8x+16

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Common factoring

To do common factoring you will first have to find a greatest common factor (GCF) in all of the terms.

for example if the expression that you have to factor is 2x+6. the GCF is 2. so you divide all the terms by 2.

2x+6

2 2

to make the factored equation, you put the factor in the front and in the bracket you put the rest. so the equation will start off as 2( )

in the bracket you have to put the left overs. so in this case 2/2 is 1 so the first term in the bracket will be x. And 6/2 is 3, so the second term in the bracket will be 3.

the factored equation will turn out to be

2(x+3)

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Simple grouping

In simple grouping you have to find the terms that are the same and put them together and put the rest together.

for example if the equation is

10(f+2)+7(f+2)

the common terms in this is f+2, so you would put them together and the rest of the terms that are left you put them together in another bracket. so the answer will be

(f+2)(10+7)

another example is 10f+20+7f+14

with this you would group the equation into two parts

(10f+20)+(7f+14)

for each bracket you have to find the GCF. so for the first bracket it will be 10 and the second bracket it will be 7. now you divide then and put the factors outside the bracket and the leftovers inside.

10(f+2)+7(f+2)

now you group them. So the final answer will be

(f+2)(10+7)

Factor Expressions of the form x2+bx+c

To factor trinomials from the form of x2+bx+c, you have to find a factor of the C that also adds to the sum of b.

for example if the equations id x2+3x+2

you have to find the factors of 2 that also add up to three, to do this you can make a chart. One column with factors, one column with product and one column with sum.

F P S so for this equation 1 and 2 will be the factors that go along with the product

1, 2 2 3 and sum.

Now to make the factored equation you turn it in to the same format as binomial products. so the answer would be

(x+1)(x+2)

Factor Expressions of the form ax2+bx+c

To factor expressions from the form of ax2+bx+c you have to find a factor that will get rid of the a and will be able to divide into b and c.

for example if your expression is 3x2+15x+18

all these terms are a product of three so we can divide all the terms by three.

3(x2+5x+6)

now you find the factors that make the product 6 and add up to 5, which is 2 and 3. so the answer for this will be

3[(x+2)(x+3)]

Word problems

A squared swimming pool has 5m added to its original area.

A) write the equation for the original swimming pool

B) find the equation for the total swimming pool

Perfect square trinomial

Perfect square trinomials terms can always be squared. the formula is (x+a)(x+a), but if there is a negative sign then one of the signs turn negative.

For example X2+36. the square root of 36 is 6. so to make this expression it will be

(x+6)(x+6) or (x+6)2

If the sign in the middle i a negative than one of the expression will have a negative sign.

(x-25). the square root of 25 is 5. so the expression will be (x+5)(x-5).


Difference of square

In difference of square the a and c are squared and the b is the the product of a and c multiplied together then multiplied by 2. if the that answer is the same as the original expression than you take the a and c and with that you get the factored expression.

For example 9x2+24x+162

the square root of 9 is three and the square root of 16 is 4. So the expression will look like

(3x)2+2()()+(4)2

in the middle of the expression with the two brackets you put the a and c in there.

(3x)2+2(3x)(4)+(4)2

then you add the middle, if the product is the same as the original term then it is a perfect square

(3x)(4)=12

2(12)=24

(3x)2+24+(4)2

so the get the factored expression for this you take the a and c vale and put it together which will be

(3x+4)2

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Solving quadratic equations

To solve quadratic equations you have to find the zeros or x-intercepts.

An example of an equation would be (x+3)(x+4)=0

you would take the first bracket and isolate x by moving the 3 to the opposite side, which will leave you with one x as -3. to find the second x- intercept you do the same thing which will give you the x-intercept of -4

Graph quadratics using x-intercepts

To graph graph by using x-intercepts you have to find other information that you need to make a parabola. Since you already have the x-intercepts you can find the axis of symmetry by adding the two zeros and dividing them by 2. To find two other points you can use the step pattern.

The quadratic formula

To find the x-intercepts from standard form equations you have to use the quadratic formula which is

x=-b+(b2-4(a)(c)

2(a)

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There are also discriminant which has the formula of (b2-4ac) If the answer to this is a negative than you won't be able to solve with the quadratic formula.

If the discriminant is greater than 0 than there will be two solutions. I there discriminant is equal to 0 than there will be 2 solutions that are the same.

Word problems
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Topic 3

Standard form to vertex form

the equation for standard form in y= ax2+bx+c and

the equation for vertex form is y=a(x-h)2+k

To change a equation from standard form to vertex form you first put a bracket in front of ax2 and end after bx.

If the equation is y= x2+4x+5

th first step would be to place the brackets

y=(x2+4x)+5

Then you divide the b value by 2, which would make it 2 and then square it which would be 4. since you cant add the 4 you will also have to subtract the four but that one will stay outside of the bracket.

y=(x2+4x+4)-4+5

inside the bracket for the h value you have to use the number you from dividing the b with two which would be 2. to find the k value you add or subtract the 2 numbers after the bracket, So the vertex form for this equation will be

y=(x+2)2+1

If there is a a value then you have to bring it outside of the bracket by dividing it with x2 and bx.

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