Question

How do u solve y=(x-3)^2-4 ???? Someoneee help meeee pleaseeee

Answer 1

do you want to find x in terms of y?

Answer 2

Yes. I keep getting y=x^2-6x+5

Answer 3

first add 4 to both sides of the formula
y + 4 = (x - 3)^2

now take square root of both sides

x-3 = +/- sqrt(y + 4)

can you continue

Answer 4

Nope. Im suppose to be solving system of equations.

Answer 5

I dont get how you would square root (y+4)

Answer 6

sqrt(x + 4) is called a radical or in UK we say Surd - it cannot be simplified anymore

Answer 7

if you are solving a system of equations where are the others?

Answer 8

square root of ( x -3)^2 is the square of ( x - 3) so by definition its square root is z - 3

Answer 9

I am suppose to be doing systems of equations. It says : On the set of axes provided,solve the following system of equations graphically for all values of X and Y. Theres two of them: y=(x-3)^2-4 and 4y+12x=20 which I already solved.

Answer 10

Im just stuck on the first equation

Answer 11

right - to solve them graphically draw them om your graph paper and where they intersect are the values of X and Y

Answer 12

Right, but when I show my work, I have to do x=-b/2a then do the X Y chart

Answer 13

you need to plot various values of X and Y for both equations

Answer 14

Exactly but I keep getting 30's and 70's as the y value

Answer 15

I get y=x^2-6X+5

Answer 16

to get values of -b/2a you need to rearrange them to the form y = ax^2 + bx + c

Answer 17

Thats what Ive been trying to seek help with!

Answer 18

sorry I'd really like to help by I have to go right now

Answer 19

Ok thnx

Answer 20

Someone help meeeeeee pleaseeeeee

Answer 21

Solve for x, then substitute the values you obtain for x and solve for y.

Answer 22

How do I rearrange the equation?

Answer 23

Its has to be in ax^2+bx+c

Answer 24

I would start with the 2nd equation, 4y + 12x = 20 and express y in terms of x like thios
4y = -12x + 20 simply by dividing both sides by 4
y=-3x + 5
Did you follow so far?

Answer 25

Yes. I got that. Thnk

Answer 26

Now, when I do the other equation I get x^2-6x+5 but Im pretty sure that is wrong.

Answer 27

Now looking at the first equation we see that y = (x-3)^2 -4 substituting or new expression for y we now have: -3x+5 = (x-3)^2 -4 do you understand that relationship?

Answer 28

Hm... Never knew we had to combine them. But yes.

Answer 29

Now go ahead and square the (x-3) and write the equations as:
-3x+5 = x^2 -6x + 9 -4 combining and simplify
x^2 - 9x = 0 do you want me to show how that comes about?

Answer 30

I will continue, factor the x^2 - 9x =0 and get x(x-9) =0 solve for x getting
x=0 and x=9 Theses are the two values of x.

Answer 31

Now substitute the value of x in one of the original equations (suggest 2nd) and solve for a corresponding y value.

Answer 32

Wouldn't it be x^2-3x+0?

Answer 33

When you simplify, I get that because you're adding 3X to -6X

Answer 34

Lets check: -3x + 5 = x^2 -6x + 9 - 4 combine similar terms on each side
-3x + 5 = x^2 - 6x + 5 now subtract 5 and add 6x to both sides getting
3x =x^2 Now subtract x^2 from both sides
-x^2 + 3x = 0 now multiply both sides by -1 (just to get a +x^2
x^2 - 3x = 0 Note I did correct an error, I was wrong about the -9.

Answer 35

Now factor x(x-3)=0
x=0
x=3 O.K so far?

Answer 36

Alrighty. I got that .

Answer 37

Yes you found my error

Answer 38

Yes

Answer 39

Lets substitute in the 2nd orig equation and find y
When x = 0 4y + 0 = 20 y=5, so one point of intersection is (0,5)
When x = 3 4y + 36 = 20, 4y = -16, y=-4 the other point is (3,-4)

Answer 40

Now the big test in order to verify those x values we substitute in the first original equation. Cross your fingers lol.

Answer 41

Right. But when I do -b/2a i get a fraction.

Answer 42

Then when I do the X Y chart it makes no sense

Answer 43

y = (0-3)^2 -4 , y=9-4=5
y= (3-3)^2 - 4 y= 0^2 -4 y=0-4 y=-4 Yeaa it is correct.

Answer 44

Yea i got that too

Answer 45

We used the substitution method in solving this, we substituted for y. What method did they tell you to uses?

Answer 46

It says to solve the following system of equations graphically for all values of x and y

Answer 47

But I didnt understand how to rearrange the first equation in order to do -b/2a then do your tables.

Answer 48

Oh, well you are going to get one curve and a line that will cross the curve at those two points.

Answer 49

The first equation y = (x-3)^2 -4 will give you the curve go ahead and work that out getting: y=x^2 -6x + 9 -4 or y = x^2 -6x +5 that is a parabola

Answer 50

Right.but I have to show -b/2a and the tables

Answer 51

Yes thats what i got but when i plug it in the calculator and see the table the y's are high

Answer 52

-b/2a = -3 I don't know why they are wanting you to plot that?????

Answer 53

Exactly. So I thought I was doing something wrong.

Answer 54

NOw do the 2nd one and you will get a linear line. where it crosses the curve are the points we have already located and is the solution of the system