Question

y=x^2-2x-15

This problem wants me to find the vertex, the x-and y-intercepts, and the symmetric point. Then I have to graph. help plz? >.<

Answer 1

I have an idea on how to go about solving this…pretty sure it involves the quadratic formula but idk so yeah. ._.

Answer 2

convert it into vertex form y = a(x-h)^2 + k where (h, k) is your vertex

Answer 3

this is usually achieved by completing the square

Answer 4

oh that thing ._. wouldn't it be like:

y+15=x^2-2x
y+15+___=x^2-2x+____
y+15+___=(x+___)^2

Answer 5

that thing?

Answer 6

EW!

Answer 7

I don't know why you moved 15 all the way to the other side of the equation, but it should work as long as you put it back where it belongs

Answer 8

take half of the x coefficient and then square it

Answer 9

y=x^2-2x-15
y = (x^2 - 2x + ______) - ____ - 15
y = (x^2 - 2x + 1) - 1 - 15
y= x^2 -2x + 1 is a complete square so you can now simplify it by factorization
then process the -1 -15 this is now your k

Answer 10

half of the x of the middle term though, because x^2 is also an x LOL @Zale101

Answer 11

so if we were to continue @Jamierox4ev3r
we end up with
y = (x-1)^2 +16

Answer 12

wait wut how?

Answer 13

hold on… ugh history homework overload in process

Answer 14

do one subject at a time

Answer 15

no i mean i just finished my history…and it kinda gave me a headache ._.

Answer 16

y+15=x^2-2x
y+15+___=x^2-2x+____
y+15+___=(x+___)^2

we can work with your way of solving it
after finding the last term to make it a perfect square, we need to add the same thing on the left hand side of the equation, correct?

Answer 17

oh wait…ha i just figured it out c;

Answer 18

ha right all the blanks would be filled with 1..

Answer 19

and you would subtract 16 from both sides, thus getting y=(x-1)^2+16

Answer 20

because the half of the middle term would be -2/2 = -1
then square
(-1)^2 = 1
so y + 15 + 1 = x^2 -2x + 1

Answer 21

yeah i see now…figuring that out was like a catharsis xDD

Answer 22

so that means that the vertex point is (-1,16) right?
and the y-intercept would be (0,15)

Answer 23

still need to find the x-intercepts tho, and that I do with the quadratic formula

Answer 24

y + 16 = (x-1)^2

then add (-16) on both sides of the equation
y + 16 + (-16) = (x-1)^2 + (-16)
y = (x-1)^2 - 16

to look for the x-intercept, your y value will be zero
so
0 = (x-1)^2 - 16

Answer 25

okay i follow so far

Answer 26

solve for x

Answer 27

wait quick question tho… would my vertex point and y-intercept be correct?

Answer 28

y intercept means your x = 0

Answer 29

right…and thats what i have up there ^ i think…*checks* yeah it is

Answer 30

oh wait, and i realized i can plug in the original equation into my calculator and check if everything is correct :P alright then

Answer 31

oh wait…its (1,-16)

Answer 32

ftw

Answer 33

did you figure out your x intercept ?

Answer 34

uh yeah..working on it

Answer 35

0 = (x-1)^2 - 16
16 = (x-1)^2
square root each side
4 = x - 1 and -4 = x -1

Answer 36

wait… oh mi gawsh thats so … ._. wow i was over thinking this

4=x-1 -4=x-1
+1 +1 +1 +1
----- -------
x=5 x=-3

Answer 37

so the x-intercepts are (5,0) and (-3,0)

Answer 38

you were probably using the quadratic formula?

Answer 39

mhm

Answer 40

try solving it using that method

Answer 41

wow holy *____fill in the blank____* wish my math teach taught us that ._. but then again, Mr. Tom Sawyer never explains anything

Answer 42

Write the quadratic polynomial on the left hand side in standard form.
Expand out terms of the left hand side:
x^2-2 x+1 = 16
Move everything to the left hand side.
Subtract 16 from both sides:
x^2-2 x-15 = 0
Using the quadratic formula, solve for x.
x = (2±sqrt((-2)^2-4 (-15)))/2 = (2±sqrt(4+60))/2 = (2±sqrt(64))/2:
x = (2+sqrt(64))/2 or x = (2-sqrt(64))/2
Simplify radicals.
sqrt(64) = sqrt(2^6) = 8 = 8:
x = (2+8)/2 or x = (2-8)/2
Evaluate (2+8)/2.
(2+8)/2 = 10/2 = 5:
x = 5 or x = (2-8)/2
Evaluate (2-8)/2.
(2-8)/2 = -6/2 = -3:
Answer: |
| x = 5 or x = -3

Answer 43

that is so much easier!!!

Answer 44

which is?

Answer 45

this way, off *mind explosion*

Answer 46

way easier ofc :P

Answer 47

both are easy

Answer 48

stupid auto correct gawsh! but thanks so much wow I'm so grateful for the help, this one makes more sense to me for some reason

Answer 49

it is just a matter of knowing which method to use given a particular situation

Answer 50

and to graph it…all i have to do is use the info (x-ints, y int., vertex)

Answer 51

exactly

Answer 52

you'll do this even in college, so you better learn it now

Answer 53

aha…EHEHEHEHEHEHEHEHE I GET IT NOW


._. okay I'm calm now I swear, but seriously thank you so much :P

Answer 54

how about this nifty solution
y=x^2-2x-15
vertex (-b/2a, y)

Answer 55

wait okay okay so uh what am i looking at here?

Answer 56

just, you know, some quick clarification >_>

Answer 57

a = 1, b = -2, c = -15
vertex (-b/2a, y)
-b/2a
-(-2)/2(1)
2/2 = 1
y = x^2 - 2x - 15, where x = 1
y = 1^2 - 2(1) -15
y = 1 - 2 - 15
y = -16

Answer 58

oh. well what do you know…i see ^_^

Answer 59

this method does not rely on completing the square, but it involves you to know that the x in vertex is solved by using the formula -b/2a
no completing the square, no moving things around and no factoring involved

Answer 60

i see…..have you ever considered being a math teacher? LOL anyhow i see how this works now, i prefer to avoid completing the square and stuff like that

Answer 61

oh yeah definitely works very well and saves you a lot of time and when it means fewer steps, fewer mistakes could be made. it does not let you off the hook solving the x and y intercepts

Answer 62

oh yeah, but that's fine :) less possibilities for mistakes are always nifty