Question

Create your own unique quadratic equation in the form y = ax2 + bx + c. Use complete sentences and show all work to determine the following:

Does the graph open up or down? How do you know? (2 points)
Explain whether the graph has a maximum or minimum point. (2 points)
Find the vertex and x-intercepts of the graph. (3 points)

Answer 1

Can you start ?

Answer 2

would this be an example? x^2 + 6x + 4

Answer 3

Sure ,why not !!
Now Can you determine the following :
Does the graph open up or down? How do you know? (2 points)
Explain whether the graph has a maximum or minimum point. (2 points)
Find the vertex and x-intercepts of the graph. (3 points)

Answer 4

the graph opens up right?

Answer 5

True ,Since (a) is +ve :)

Answer 6

the graph has a minimum point?

Answer 7

Since it opens upwards, it must have a minimum value.
Graph it and find it out :)

Answer 8

yea its minimum :) what is the formula for vertex?

Answer 9

@Eyad

Answer 10

Since this is a quadratic equation , Vertex = b/2a

Answer 11

you chose
would this be an example? x^2 + 6x + 4

this is a valid answer, but it is "hard" to find its x-intercepts (you have to use the quadratic formula because it does not factor nicely.

If you start with for example, (x+1)(x-2) and multiply it out you get a nice quadratic

Answer 12

:o i want the easiest way lol so let me do (x + 1)(x - 2)

Answer 13

how do i solve that ^^^^^^^^^^^^^^^ ???

Answer 14

use FOIL to multiply it out so it is in "standard form"
can you mutliply it out ?

Answer 15

x^2 .... ummmm

Answer 16

FOIL is short for First x*x
outer -2*x
inner 1*x
last -2*1
you get x^2 -2x + x -2
combine the -2x+1x to get -1x

finally: y= x^2 -x -2

remember it factors into (x+1)(x-2) and you can solve for
(x+1)(x-2)=0 to find the x-intercepts

Answer 17

this makes a smile because the "a" is positive (a is the number in front of x^2, a is +1)

Answer 18

so it still opens up and it still has a minimum point

Answer 19

@phi how do i find the vertex?

Answer 20

@phi ;/

Answer 21

your quadratic is
y= x^2 -x -2

this is in the form a x^2 + bx +c
where a =1, b= -1 and c= -2
the vertex occurs at x = -b/(2a)

replace the letters with the numbers. what do you get for x ?

Answer 22

-1/2*1 ????

Answer 23

- 1/2 ??

Answer 24

what is -b ? it is - (-1) and the 2 minus signs make a +
try again

Answer 25

so it would be - (-1)/2*1 ?

Answer 26

yes though you should write it as
- (-1)/(2*1)
with the 2*1 in parens...
that simplifies to what ?

Answer 27

1/2

Answer 28

right?

Answer 29

yes, x= 1/2
now find the y value that goes with this x value
replace x with 1/2 in
y =x^2 -x -2

you can use a calculator if you have one

Answer 30

wait.. the vertex was x = 1/2 ? or we r in the process ?

Answer 31

yes the vertex has an x value of 1/2.
I assume they want the (x,y) pair for the vertex, so you should find the y value also

Answer 32

ok.. y = -9/4

Answer 33

yes, so the vertex is at
(0.5, -2.25) in decimals or
(1/2, -9/4) as fractions

finally, the x-intercepts happen when
(x+1)(x-2) =0

this means either x+1 =0 or x-2=0
solve both equations to find the 2 x-intercepts

Answer 34

do u normally do it fractions or decimals?

Answer 35

as a check, we can use wolfram
http://www.wolframalpha.com/input/?i=plot+y%3D%28x%2B1%29%28x-2%29+

click on the properties button in the bottom panel
to see the vertex

fractions are fine. In the real world, we use decimals, but in school fractions are easy to read

Answer 36

to solve x+1=0 you could add -1 to both sides:
x + 1 -1 = 0 - 1
and simplify to get
x = -1

I assume you can solve x-2 = 0 ?

Answer 37

now we r gonna find the x- intercepts right?

Answer 38

@phi

Answer 39

yes, find the x-intercepts. but because we started with
y= (x+1)(x-2)
you can find them by solving
(x+1)(x-2) = 0 (x intercepts are the x values that make y = 0)



to solve x+1=0 you could add -1 to both sides:
x + 1 -1 = 0 - 1
and simplify to get
x = -1

I assume you can solve x-2 = 0 ?

Answer 40

x = 2

Answer 41

@phi

Answer 42

I think you have answered all the questions.

the x-intercepts can also be written as (x,y) pairs: (-1,0) and (2,0)
but it is generally understood that the y value of an x-intercept is 0

Answer 43

so the x intercepts are (-1, 2)

Answer 44

thank you!!

Answer 45

I would not put them in parens, because that might be confused with the point (-1,2)

best to say (-1,0) and (+2,0)

Answer 46

ok :)

Answer 47

let y=2x^2+4x+3
=2(x^2+2x)+3
=2(x^2+2x+1-1)+3
=2(x+1)^2 -2+3
=2(x+1)^2+1
or (y-1)=2(x+1)^2
or (x+1)^2=1/2 (y-1)
It is an upper parabola with vertex at (-1,1) &x-intercept of vertex=-1