Quadratics

Parabolas

Parabolas is one of three topics in Quadratics. Linear systems is graphing a straight line, but parabolas are a curve. It is like kicking a football into the air - it will go up in a arc and come down again. Here, you will learn the features of parabolas and how to graph from vertex and factored form.
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  • parabolas can open up or down
  • the zero of a parabola is where the graph crosses the x-axis
  • "zeros" can also be called "x-intercepts" or "roots"
  • the axis of symmetry divides the parabola into two equal halves
  • the vertex of a parabola is the point where the axis of symmetry and the parabola meets. It is the point where the parabola is at its maximum or minimum value
  • minimum = parabola opens up
  • maximum = parabola opens down
  • the optimal value is the value of the y co-ordinate of the vertex
  • the y-intercept of a parabola is where the graph crosses the y-axis
Analyzing Quadratics

Understand the base graph, what makes the graph curve, and the step pattern.

Base graph

y=x2
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The base graph starts on (0,0). As you can see the step pattern is 1, 4, 9... (based on the chart), which means move on to the right & go up one, move two to the right & move up four, move to the right three & move up four, and so one. Also do this to the left side and create the parabola! (however, if it is negative, just move down instead of up, you will learn more about reflecting upon th x-axis later),

Vertex Form

y=a(x-h)2+k

What Effects Does Each Part of the Vertex Equation Have On The Graph?

Vertex equation: y=a(x-h)2 +k

a = stretch or compression

  • stretch is a whole number
  • compression is a decimal/fraction that is less than one
  • it either opens up (positive) or down (negative)


h = x value of vertex

  • moves left and right


k = y value of vertex

  • moves up and down
  • optimal value


for example: y=(x-4)2+6

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As you can see, in the equation y=(x-4)2+6, it moved 4 to the right and 6 up from the base graph, and the vertex was at (4,6). This vertex form equation did not specify an value, thus equalling 1. However, if 'a' is not 1 or negative 1, it will specify on the equation, and you would have to multiply your 'a' with the step pattern. For example: y=2(x-4)2+6
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Finding the Equation Given The Vertex

If you are given the vertex points and another point, you can figure out what the 'a' value is

Try this out!

State each term for the parabola on the right:

Vertex -

Direction of opening -

Max or min value -

Axis of symmetry -

Quadratic relations equation in vertex form -



Ready for the answer?

Vertex - (2,1)

Direction of opening - up

Max or min value - minimum, y=1

Axis of symmetry - x=2

Quadratic relations equation in vertex form - y=(x-2)2+1

Try this out!

Write an equation of a parabola that has a vertex at (2,1) and goes through the point

(3, -7).



Ready for the answer?

y=a(x-h)2+k

y=a(x-2)2+1

-7=a(3-2)2+1

-7=a(1)2+1

-7=1a+1

-8=1a

-8=a


Therefore, the equation is: y=-8(x-2)2+1

Try this out!

graph the following equation:

y=2(x-4)2+6


Ready for the answer?

(graph attached to the right)

Try this out!

A football kicked into the air follows a parabolic path described by the equation:

h=-2(t-3)2+19 h is verticle height in meters, and t is time in the air in seconds.


a) when does the football reach its maximum height?

b) what is the maximum height reached?

c) what is the height of the ball when punted?


Ready for the answer?

a) 3 seconds

b) 19 meters

c) ball was punted at 1 meters high

t=0


y=-2(t-3)2+19

y=-2(0-3)2+19

y=-2(3)2+19

y=-2(9)+19

y=-18+19

y=1

Graphing From Factored Form

Showing you all the steps to graph from factored form -> y=a(x-r)(x-s)

Try this out!

GRAPH:

y= -1/2 (x+3)(x-6)



Ready for the answer?

zeros: (-3,0) (6,0)

axis of symmetry: 1.5

optimal value: 10.25

REFLECTION

Factoring!

There are 6 types of factoring:


1. common factoring

2. simple trinomial

3. complex trinomial

4. factoring by grouping

5. perfect squares

6. difference of squares

1. Common Factoring

instead of multiplying polynomials, we are going to factor polynomials

Factors - numbers that are multiplied to get a certain number (common factors of 6 is 1, 2, 3, 6 because 1x6=6 & 2x3=6)


say you had to common factor the equation 2x+4. 2 is common with both values. so, you would divide 2 out from that equation and put what you factored out (2) in the middle of your new equation along with the divided factor; so the new equation would look like =2(x+4)


another example is x2+6x. x is common with both values. so, you would divide x out from that equation and put what you factored out (x) in the middle of your new equation along with the divided factor; so the new equation would look like =x(x+6)


one last one would be 2x3+8x2. if you are dealing with powers in both terms, take the lowest power so factor out. so, 2x2 is common with both values. so, you would divide 2x2 out from that equation and put what you factored out (2x2) in the middle of your new equation along with the divided factor; so the new equation would look like =2x2(x+4)

2. Factoring Simple Trinomials

When factoring it, we are trying to break it up into two things that multiply up to make it (two bracket). You are looking for two numbers that multiply to get the last term, and are the sum of the second term. You may want to do some rough work on the side and list all the possible numbers that would be the product of the last term. Then, look at which ones add up to your second term. After you figure out what numbers to use, use the guess and check method in your two brackets. To check if your answer is right, just expand to get the original equation.

Try this out!

a) p2+7p+12

b) r2-10r+16

c) w2+2w-48

d) x2-3x-18


ready for the answer?

a) p2+7p+12 =(p+3) (p+4)

b) r2-10r+16 =(r-2) (r-8)

c) w2+2w-48 =(w-6) (w+8)

d) x2-3x-18 =(x+3) (x-6)

3. Factoring Complex Trinomials

a complex trinomial is a trinomial which starts with a coefficient thats not equal to 1.

the first thing to look for in a complex trinomial is if you can change it from a complex to a simple trinomial. this can be done by common factoring it. for example: 3x2-6x+9. it could be common factored by 3 and be changed into =3(x2-2x+3) and thus changed into a simple trinomial.

but you cant always do that, and if you cant then you would have to do another method; which is guess and check. on the side, list all the possible numbers that would be the product of the first term and the last term. then plug them in in two brackets and check your answer by expanding & collecting like terms to get the original equation.

Try this out!

a) 4x2+25x-21

b) 2x2+11x+15


ready for the answer?

a) 4x2+25x-21 =(4x-3)(x+7)

b) 2x2+11x+15 =(2x+5)(x+3)

4. Perfect Squares

perfect squares are one of the two special factoring cases.

  • you know its a perfect square if you can square root a & c
  • check by multiplying answer by 2, which should equal b

Try this out!

x2+8x+16


ready for the answer?

x2+8x+16

=(x+4)(x+4)

=(x+4)2

check:

(x+4)(2)

=8x

5. Difference of Squares

difference of squares is the other special factoring case.

  • you know its a difference of squares if it is 2 terms, and you can square root both
  • the plus and minus is what makes the minus in the first equation

Try this out!

x2-9


ready for the answer?

x2-9

=(x+3)(x-3)

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need extra practice? try this out!

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Reflection

Completing The Square

There are 4 steps to completing the square. An important formula to remember is (b/2)squared. Steps:

1. remove the common factor from the x2 and the x-term coefficient

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

3. Group the three terms that form the perfect square. (move the subtracted value outside the bracket by multiplying it by the common factor first)

4. factor the perfect square and collect like terms

(example on the picture to the right)

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Solving quadratic equations from standard form (using the quadratic formula)

We already seen some stuff that we could do with standard form, but it involves converting the equation to vertex form. the quadratic formula would allow you to work straight from standard form.

if you have a standard form quadratic equation that equals zero (has an a, b, & c value) (0=ax2+bx+c), then the formula looks like x=negative b plus/minus square root of b squared minus 4ac, all divided my 2a. the plus/minus will give the two solutions.

for example, find the x-intercepts of y=2x2+3x-5

the a value is 2 from 2x2

the b value is 3 from 3x

the c value is -5 from -5

now plug in all the values into the quadratic formula!

(work shown on the picture to the right)

after you get to the part where you square root your number, the plus and minus indicates that there will be 2 solutions.

discriminant - the value in the square root

  • if you have a negative discriminant, it is not possible because you cannot square root a negative number. therefore there would be no solutions (x-intercepts).
  • if you have a positive discriminant, there would be two solutions (x-intercepts).
  • if you ave a discriminant that is equal to zero, there is only one solution (x-intercept).


Try this out! (word problems)