Question

ompare and contrast the two quadratic equations below. In order to receive full credit, use complete sentences to describe the following:
The direction each parabola opens
The vertex of each parabola
y = x2 - 2x
y = -2x2 + 4x - 3

Answer 1

@ParthKohli

Answer 2

@thomaster

Answer 3

YES HELP ME!!!

Answer 4

...

Answer 5

You can tell the direction by looking at a.
The equation for this is:
Ax + By = C
It has been reduced to:
By = Ax - c

Answer 6

If A is negative, the parabola looks like a frown, or opens downwards.

Answer 7

no i keep getting (1,-1) for my vortex!!

Answer 8

i know how to do it :3

Answer 9

Let me check. :)

Answer 10

k

Answer 11

cause its stupid, both of em is (1,-1) for me ;P

Answer 12

One second..

Answer 13

k

Answer 14

f(x) = x^2 - 2x
f(1) = 1^2 - 2(1)
f(1) = 1 - 2
f(1) = -1
So I know that that's an ordered pair (maybe not the vertex)

f(1) = -2x^2 + 4x - 3
f(1) = -2(1)^2 + 4(1) - 3
f(1) = -2 + 4 - 3
f(1) = -1
So I know that's an ordered pair.

Answer 15

sry brb

Answer 16

I'm gonnna use the quadratic formula for this.

Answer 17

1. \(ax^2 + bx + c\) is the general equation of a parabola. If \(a > 0\), then the parabola is an upwards U. If \(a < 0\), then the parabola is a downwards U, which kinda looks like \(\cap\).

2. The vertex of a parabola is given by \(\left(-\dfrac{b}{2a}, f\left(- \dfrac{b}{2a}\right)\right)\). Do you know why that is?

Answer 18

kk

Answer 19

back

Answer 20

y = x^2 - 2x
\[x = \frac{ -b \pm \sqrt{b ^{2} - 4ac }}{ 2a }\]b = -2
a = 1
c = 0
b = -2\[x = \frac{ 2 \pm \sqrt{-2 ^{2} - 4(1)(0)} }{ 2(1) }\]
\[x = \frac{ 2 \pm \sqrt{4 } }{ 2 }\]
\[x = \frac{ 2 \pm 2 }{ 2 }\]
\[x = \frac{ 2 }{ 2 } \pm \frac{ 2 }{ 2}\]
\[x = 1 \pm 1\]
\[x = 2, x = 0\]
Our x vertex will be between these two points. (1, ?)

Answer 21

i got -1

Answer 22

TOLD YOU THIS

Answer 23

You got -1 for y, now we solve for y.

Answer 24

Then I can see what the vertex is.

Answer 25

ITS NOT THAT I DONT KNOW HOW TO DO THIS! ITS THAT I KEEP GETTING 1 AND -1!!!

Answer 26

I understand that, but what makes you think you're wrong?

Answer 27

why! why i keep getting -1!!

Answer 28

Because it could be the correct answer.

Answer 29

oh its that the other vortex is also 1 and -1

Answer 30

Let me check, calm down please.

Answer 31

thats why

Answer 32

y = x^2 - 2(1)
y = 1^2 - 2(1)
y = 1 - 2
y = -1
Vertex for y = x^2 - 2x
(1,-1)

Answer 33

oh so you got that too?

Answer 34

Okay, so you are correct for the first equation. I'll check the second if you don't get upset.

Answer 35

*heavy breathing like a psycho

Answer 36

*hands Epicteatime a chill pill*

Answer 37

@jcpd910

Here's the easy way out: https://www.desmos.com/calculator

Answer 38

*receives

Answer 39

Err @Epicteatime

Answer 40

NO

Answer 41

iGreen, what about when he's doing a written exam and he doesn't have the option to use a quad calc?

Answer 42

lol

Answer 43

He should know how to use this, so calculators are not the option. If they were, nobody would have to do real math.

Answer 44

Epic, want me to check your work?

Answer 45

Actually, they would. Graphing calculators only make it easier to graph things.

Answer 46

for the second eqaution?

Answer 47

?

Answer 48

yep

Answer 49

iGreen, it's nice to know how to do this. -.-

Answer 50

yes please

Answer 51

Right, going in! *puts on war paint*

Answer 52

Lol..I already know the long way of doing it..so I just use the Calculator..it's just faster.

Answer 53

lol

Answer 54

y = -2x^2 + 4x - 3
\[x = \frac{ -b \pm \sqrt{b ^{2} - 4ac }}{ 2a }\]
a = -2
b = 4
c = -3
\[x = \frac{ -4 \pm \sqrt{4^{2} - 4(-2)(-3) }}{ 2(-2) }\]
\[x = \frac{ -b \pm \sqrt{16 - 24 }}{ -4 }\]
\[x = \frac{ -4 \pm \sqrt{-8}}{ -4 }\]

hmmm looks like we're gonna have to use complex solutions, which I think means we aren't gonna have (1,-1)

Answer 55

WOOOOO

Answer 56

what the crap is this

Answer 57

Wait though, a complex solution still is a solution, to the point where we can solve it. problem with me is that I'm not good at complex solutions. O.o

Answer 58

lol

Answer 59

so...

Answer 60

Have you been taught complex solutions yet?

Answer 61

nope

Answer 62

i 'started' learning this like a few days ago...