My math = fail. Help? XD

[Kage]

My math = fail. Help? XD

You know I'm desperate when i go to forums to ask for someone to solve things..

I'm out for a reason. XD This is it. I just can't seem to do this at all.

I'm supposed to find quadratic functions (be it in standard, intercept, or vertex forms) that satisfy the give information. Anyone wanna help? XD Or give me the answers? XD

1)one of the x-intercept -4, y-intercept 24, maximum value 27
I gather that a is negative. C=24. I have points (0,24) and (-4,0) and (x,27) since if it's the max, you'd also gather that that is the vertex. I've tried solving it using the intercept form... variables got in the way and confused me. Same thing with vertex form. I'm hoping if I can understand this problem I can do the rest.

Help? Please? THANKS~

 

[Tcatomon]

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*head explodes*

 

[Airdra]

I used to know how to do this. Years ago, before I went to college and selected the major that would have me take the least possible amount of math. >_> It's like this close, but I just can't remember.

Does this help any? If that doesn't help, I'm working on a bit of a math review for my GRE, and I can make an attempt at working the problem a little later tomorrow when I'm actually awake if the GRE review book I'm using covers it. ^^;;

 

[Kage]

My algebra 2 obviously is not as good as the A- in the Alg.2/Trig honors class made it... @_@ I knew what my teacher wanted me to know and that was enough. XD Which, obviously did not extend to this.

 

[Elecmon]

The summer vacation has fried my brain. I'm going into Calculus and from what you said I can't figure it, but that may be because I haven't written it down. Are you trying to find the equation of the graph?

 

[Kage]

I can pretty much do what that page says. I just can't find the functions with these "given"s that are supposed to satisfy the functions.

 

[Kage]

The summer vacation has fried my brain. I'm going into Calculus and from what you said I can't figure it, but that may be because I haven't written it down. Are you trying to find the equation of the graph?

I just got out of Pre-calc... and yah... I'm trying to find the function which would be like ax^2+bx+c or in the intercept form... or the vertex form.

 

[RamenRanger]

only if you got this two months ago the summer fried my brain in this stuff

ill see if i still have my notes some where tomorrow morning...but the chance of me finding it are slim.

 

[Kage]

It's ok. If someone can explain it to me, it'd be great. I can fail homework... I don't really want to fail the exam later. @_@ I don't want more of this freaking class.

I'm taking like 4+ hours each day of classes because of this shit. ;-; I'm exhausted.

 

[Tcatomon]

And that's why I took pre-calc and Physics twice >_>;

 

[GSR]

...

Sheeeeeeeeeaaaaaaat. I'm this close to remembering how to solve these, but I haven't taken Algebra for over a year now, so my brain is dead in that respect. D:

 

[celestial_sacred]

Do you still need the answer?? 'Cause I think I got it.. Haha.

Intercept-x = -4
Intercept-y = 24
Maximum-y = 27
Maximum graph, a < 0

So, there are two possibilities.
View attachment 626

y = ax[sup]2[/sup] + bx + c
Since intercept-y = 24, c = 24
y = ax[sup]2[/sup] + bx + 24 ----- (1)

From (1), when x = -4, y = 0,
0 = (-4)[sup]2[/sup]a + (-4)b + 24
= 16a - 4b + 24
4b = 16a + 24
b = (16a + 24) / 4
b = 4a + 6 ----- (2)

From (1),
y = ax[sup]2[/sup] + bx + 24
= a (x[sup]2[/sup] + (b/a)x) + 24
= a (x[sup]2[/sup] + (b/a)x + (b/2a)[sup]2[/sup] - (b/2a)[sup]2[/sup]) + 24
= a (x + (b/2a))[sup]2[/sup] - a(b/2a)[sup]2[/sup] + 24
= a (x + (b/2a))[sup]2[/sup] - a(b[sup]2[/sup]/4a[sup]2[/sup]) + 24
= a (x + (b/2a))[sup]2[/sup] - b[sup]2[/sup]/4a + 24
____________________^
|
I think you know this part gives the max-y/min-y value

Since maximum-y = 27,
-b[sup]2[/sup]/4a + 24 = 27
-b[sup]2[/sup]/4a = 3
Substitute (2) into the equation,
-(4a+6)[sup]2[/sup]/4a = 3
-(16a[sup]2[/sup] + 48a + 36) = 12a
16a[sup]2[/sup] + 48a + 36 = -12a
16a[sup]2[/sup] + 60a + 36 = 0
4a[sup]2[/sup] + 15a + 9 = 0
:
: (okay. i'm lazy here)
a = -3/4 or -3

Substitute a = -3/4 into (2),
:
:
b = 3
Substitute a = -3 into (2),
:
:
b = -6

Substitute a = -3/4, b = 3 into (1),
y = -3/4x[sup]2[/sup] + 3x + 24

Sutstitute a = -3, b = -6 into (1),
y = -3x[sup]2[/sup] - 6x + 24


So, you get two quadratic functions.. haha. Should be correct since the a is less than 0 and if you use the "complete the square" method, you'll get max-y = 27. Well, and as you know, I did this according to what I've learnt..
Sorry for the extra LONG working. I could skip a lot but I was afraid you might not get it..

What's vertex by the way? Ahaha.. I used to learn Maths in my national language but it just changed back to English a few years ago. But I started learning in English this year..pre-u..so, I'm a little stupid in the terms part..!

 

[RamenRanger]

/me brain implodes.

 

[Ainz]

If you wanted to find the two possible x-coordinates of the maximum, differentiate both the solutions and set them both equal to zero, as the gradient at the maximum is zero

y = -1/2x[sup]2[/sup]+ 4x + 24 - dy/dx = 4-x

y = -13/4x[sup]2[/sup] - 7x + 24 - dy/dx = -13/2x-7

x=4, -14/13

(4, 27) and (-14/13, 27) both clearly evident on the two above graphs.

 

[Skeith]

He has the maximum, 1 of 2 x-intercepts, and the y-intercept. He doesn't have the graph yet, so the search for the x-value isn't the problem.

The best way would probably to put it into
y=a(x+(b/2a))^2 - ((b^2)/4a)+c form, which celestial_sacred did.

The only change is that I would make -((b^2)/4a) to a = -(b^2)/12, and sub the other way, if only to make it slightly less complicated.

Let it be known that ( -(b/2a), f(-(b/2a)) ) is the value of the vertex in quadratics. In this case, f(-(b/2a)) would be ((b^2)/4a)+c.

I could've sworn there was some different way... Aaargh, it was so long ago, like 2 years or something.

Vertex is the maximum/minimum point in a quadratic, if I remember correctly...

 

[celestial_sacred]

Vertex is the maximum/minimum point in a quadratic, if I remember correctly...

Ah...~~ Thanks!

 

[nezucho]

math makes my head hurt *runs*

 

[SSJ Jup81]

I despise this type of math!! Why the hell do we actually need this crap!! It doesn't help me with my day-to-day functions, like grocery shopping or bill paying. I don't feel that we should be forced to learn this mess. Pre-calc is as high as I got and I barely passed that since I didn't understand it.

 

[Kage]

Yeah well I wanna be an electrical engineer or a computer engineer...

@_@ ANYWAYS, there's been a few days but I still feel as if my brain exploded. So I'll wait a few more days to see if I TRULY understand. XD

 

[SSJ Jup81]

Yeah well I wanna be an electrical engineer or a computer engineer...

That's different. If one is going into a field that requires difficult math, then no problem, but why do we all have to suffer if we have no intention or desire to pursue careers where that type of difficult (and confusing) math is required? I never understood why it's mandatory to learn certain types of math. The only type of math that should be mandatory is basic because we use it for everyday functions.