# Thread: Help me with graphing parabolas

1. ## Help me with graphing parabolas

failed a unit test retaking it but I don't know how to do some of the questions please help explain..
y = ( x + 2)^2 -3 its not in y = a ( x - h)^2 + k so I don't use h,k for my min or max value and axis of symettry point on the graph. {ps question if I did if the question represented itself as y = 3/4 ( 2 - 2)^2 + 3 , would my x,y be negative 2 and 3 (-2,3) ... or is it supposed to become a positive (2,3)
so y = ( x + 2)^2 -3 the equasion is also not in y=ax^2 + c so I don't use x= -b/2(a) to find x either...I'v looked through book examples .. how do I solve this? Also Its algebra 2 translating parabolas nothing too fancy please..I messed up on my test this way because I saw problems like these and saw no a and pretty soon I dropped a completly on all my graphing problems when using the vertex form. y = a ( x - h)^2 + k How am I supposed to solve this for my test if y = ( x + 2)^2 -3 has no a ?
Thanks
edit: please tell me if I can make this any more clear and I don't need to see the graph just how to get up to that point

2. Originally Posted by D@nny
failed a unit test retaking it but I don't know how to do some of the questions please help explain..
y = ( x + 2)^2 -3 its not in y = a ( x - h)^2 + k so I don't use h,k for my min or max value and axis of symettry point on the graph. {ps question if I did if the question represented itself as y = 3/4 ( 2 - 2)^2 + 3 , would my x,y be negative 2 and 3 (-2,3) ... or is it supposed to become a positive (2,3)
so y = ( x + 2)^2 -3 the equasion is also not in y=ax^2 + c so I don't use x= -b/2(a) to find x either...I'v looked through book examples .. how do I solve this? Also Its algebra 2 translating parabolas nothing too fancy please..I messed up on my test this way because I saw problems like these and saw no a and pretty soon I dropped a completly on all my graphing problems when using the vertex form. y = a ( x - h)^2 + k How am I supposed to solve this for my test if y = ( x + 2)^2 -3 has no a ?
Thanks
edit: please tell me if I can make this any more clear and I don't need to see the graph just how to get up to that point
In this case you would treat a as one, that is, take a=1 and continue as you would normally.

So as usual, since the x^2 will end up being positive, we know we will have a minimum value, and it is the same (h,k) which is (-2,-3) in this case. We then find the x and y intercepts. For x-intercept, we set y=0. That is, 0=(x+2)^2 - 3
=> x^2 + 4x + 4 -3 = 0
=> x^2 +4x +1 = 0
We proceed by the quadratic formula, using a=1, b=4, c=1
We get, x = [-4 +- squareroot(4^2 - 4(1)(1))]/2(1)
simplifying we get x= -2-3^(1/2) and -2+3^(1/2) --> the x-intercepts.

Now the y-intercept(s), we set x=0.
So y=(0+2)^2 -3 = 4-3=1

This is all we need for a parabola. So we end up with:

[IMG]file:///C:/DOCUME%7E1/Jhevon/LOCALS%7E1/Temp/moz-screenshot.jpg[/IMG]

3. That graph doesn't look as clear as I expected it to be. Can you see all the points clearly? The min is (-2,-3), the x-intercepts are (-2-3^(1/2) , 0) and (-2+3^(1/2) , 0), the y-intercept is (0,1). You don't have to be overly precise trying to mark off a scale on the axis, just draw a graph resembling the one you see and label the intercepts and critical points (maximums and minimums)

4. Originally Posted by J
=> x^2 + 4x + 4 -3 = 0
=> x^2 +4x +1 = 0
We proceed by the quadratic formula, using a=1, b=4, c=1
We get, x = [-4 +- squareroot(4^2 - 4(1)(1))
/2(1)
simplifying we get x= -2-3^(1/2) and -2+3^(1/2) --> the x-intercepts.

Now the y-intercept(s), we set x=0.
So y=(0+2)^2 -3 = 4-3=1

This is all we need for a parabola. So we end up with:
You lost me at ==> please explain how you got x^2 + 4x + 4 -3 = 0

5. I expanded the perfect square, (x+2)^2 = x^2 + 4x + 4, then I subtracted the -3 as the formula directed

6. Originally Posted by Jhevon
I expanded the perfect square, (x+2)^2 = x^2 + 4x + 4, then I subtracted the -3 as the formula directed
Are you using the distributive property? wouldn't it be x^2 + 4 - 3 ?
Why + 4x + 4?

7. Well, yeah, it's using the distributive property in a weird way. REMEMBER, THE ^2 IS SQUARING, NOT MULTIPLYING. There are two ways to do this. The first one is what I usually do.

We have (x+2)^2. The expansion is the square of the first term + 2*the product of both terms + the square of the second term. So (x+2)^2 = x^2 + 2*(x*2) + 2^2 = x^2 + 4x +4.

The other way is using the distributive property directly. Rewrite (x+2)^2 as (x+2)(x+2) and distribute. Take the x in the first group and multiply everything in the second group, then take the 2 in the first group and multiply everything in the second group. So (x+2)(x+2)=x^2 + 2x + 2x + 2*2 = x^2 + 4x + 4. I never do this though, it takes to much time and writing. Note, the first method only works for binomial squares. If you had for instance (x+2)^3, you would need something different.

8. Thanks..yeah the second method seems easier for me to follow.
Edit: so if I make a = 1 and solve it as I would using the equation of a parabola I would get the same answer?

9. Yeah. That is what I did afterall. Maybe your professor taught you to lay out the problem a different way from what I did, but the underlying principle is the same. Using a=1 in whatever method you use should yield the same result. So your good now right?

10. am I on the right track ... y = 1(a) (x+2)^2 - 3 my vertex point is -2,-3 [you flip the sign for h because in the formula its -h right?]
But then I plug in -1 for x and get y = -2. I plug in 0 and I get positive 1. I think i'm making this too simple now. If I plugged in 0 my y intercept should be the same thing you got right?
Your graph is too complicated. I'v never seen coordinates that way before.
heres my graph http://img394.imageshack.us/img394/8103/grpjo4.png (-3 , -2)(-1 , -2)

11. Originally Posted by D@nny
am I on the right track ... y = 1(a) (x+2)^2 - 3 my vertex point is -2,-3 [you flip the sign for h because in the formula its -h right?]
But then I plug in -1 for x and get y = -2. I plug in 0 and I get positive 1. I think i'm making this too simple now. If I plugged in 0 my y intercept should be the same thing you got right?
Your graph is too complicated. I'v never seen coordinates that way before.
Why did you plug in -1 for x? I don't follow your reasoning. Plugging in 0 does result in positive 1, which is the y-intercept. You were correct in finding the minimum, now your next objective is to find the intercepts. For the x-intercept you plug in y=0, ALWAYS, and for the y-intercept, you plug in x=0, ALWAYS. If you prefer, you can write the x-intercepts as decimal expansions, but it makes no difference: (-3.732, 0) and (-0.268, 0).

It might be a while before my next reply, about half an hour or so, so if you have more questions, don't hold your breath, I gotta go take care of something.

12. Originally Posted by Jhevon
Why did you plug in -1 for x? I don't follow your reasoning. Plugging in 0 does result in positive 1, which is the y-intercept. You were correct in finding the minimum, now your next objective is to find the intercepts. For the x-intercept you plug in y=0, ALWAYS, and for the y-intercept, you plug in x=0, ALWAYS. If you prefer, you can write the x-intercepts as decimal expansions, but it makes no difference: (-3.732, 0) and (-0.268, 0).

It might be a while before my next reply, about half an hour or so, so if you have more questions, don't hold your breath, I gotta go take care of something.
I'm not using a graphing calculator. But why are you getting decimals?? I think I'v solved this problem now. (-3.732, 0) and (-0.268, 0) . Cool math .com - Online Graphing Calculator - Graph It! . Yeah those look right..my answers are right also . I dont have a graphing calculator yet..were working on finding these things without going into decimals

13. The way I had it before had no decimals, I just wrote decimals cause you said they look weird. -2-3^(1/2) and -2+3^(1/2) are exact values.

14. yeah your answers are also parabolas of -2. Basically that equasion is plug and chug. It doesn't really matter what your coordinates are just as long as their parabolas. And I think it's much easier just to go up/down by integers when graphing so you don't have to get into decimals..

15. Originally Posted by D@nny
yeah your answers are also parabolas of -2. Basically that equasion is plug and chug. It doesn't really matter what your coordinates are just as long as their parabolas. And I think it's much easier just to go up/down by integers when graphing so you don't have to get into decimals..
If you don't want to use decimals, then use the exact values. Using integers would be considered wrong. For instance, if you used integers you would have the x-intercepts being (-4,0) and (0,0). The (0,0) says the graph goes through the origin, which we know is not the case. Don't worry about how it looks, just make sure your calculations are correct. If you get roots in your answer for an intercept, thats fine, just write it on the graph in it's relative position. For instance, -2-3^(1/2) is a little less than -2. So you just pick some arbitrary point between -2 and -3 on the x-axis and label it -2-3^(1/2), and be done with it.

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