Calculus Help?

[Yusuke Urameshi]

Here's the problem. I know you guys are smart enough to figure this out, because I'm not 100% sure on how to do it. Any legitimate attempts will be repped, if I can rep you.

Redo problem 27 in section 4.1 of your textbook (page 178), but find the parabola whose minimum is at (10,2) rather than the point given in the book.

The parabola's equation is y = x^2 + ax + b, where
a =
and b =

 

[xxsasukex]

I'll Be back In A Few with the answer, i Need to find a pen

 

[Alfred Pennyworth]

what is a and b? what textbook? what answer?

 

[Yusuke Urameshi]

I'll Be back In A Few with the answer, i Need to find a pen

lol ok

what is a and b? what textbook? what answer?

You have to find a and b. The textbook isn't relevant. I'm trying to find the answer. lol

 

[Titania Erza]

That ain't even worth calling calculus XD

y = x^2 + ax + b so the derivative is equal to: 2x+a

2x+a=0 for x=10
2*10+a=0
a=-20

Given (10,2) so 2 = 10^2-20(10)+b
2 = 100-200+b
b=2-100+200=102

 

[Sir Blades]

That ain't even worth calling calculus XD

y = x^2 + ax + b so the derivative is equal to: 2x+a

2x+a=0 for x=10
2*10+a=0
a=-20

Given (10,2) so 2 = 10^2-20(10)+b
2 = 100-200+b
b=2-100+200=102

you got it......

 

[fiend]

differentiate the equation y = x^2 + ax + b which gives us y'= 2x+a

now put y'=2x+a =0

i.e 2x+a=0 substitute the value of x=10 so a=-20 because minima criteria dy/dx=0

now since (10,2) lies on parabola

so, 2=10^2 +ax +b

i.e 2=100-200+b

b=102

Edit:- Someone already solved it. :O

 

[Yusuke Urameshi]

That ain't even worth calling calculus XD

y = x^2 + ax + b so the derivative is equal to: 2x+a

2x+a=0 for x=10
2*10+a=0
a=-20

Given (10,2) so 2 = 10^2-20(10)+b
2 = 100-200+b
b=2-100+200=102

Yeah, it's prolly more like pre-calculus. This curriculum sucks. The prof. did a whole chapter on derivatives without even showing the easy way to do it. I took calculus in high school so I was completely fine, but I felt sorry for all the plebeians. lol

 

[Titania Erza]

Yeah, it's prolly more like pre-calculus. This curriculum sucks. The prof. did a whole chapter on derivatives without even showing the easy way to do it. I took calculus in high school so I was completely fine, but I felt sorry for all the plebeians. lol

No problem and yep this would be labeled as one of the easiest parts of pre-calculus. If you ever need any help with maths, you can send me a PM.

 

[Alfred Pennyworth]

oh dayum I see it now..

welp , since numerous solutions with derivatives are already available , I'll show you one without any differentiation then..


rewrite x^2+ax+b into (x+(a/2))^2+(b-(a^2/4)) by completing perfect square

we know that (10+(a/2))^2=0 so we know a=-20

and (b-(a^2/4))=2 makes b=102

 

[Dãrth1]

Just calculate it with X = 10 and Y = 2 and you'll do it

 

[YellowFang]

That ain't even worth calling calculus XD

y = x^2 + ax + b so the derivative is equal to: 2x+a

2x+a=0 for x=10
2*10+a=0
a=-20

Given (10,2) so 2 = 10^2-20(10)+b
2 = 100-200+b
b=2-100+200=102

My my... That's quick.

 

[Uzumaki Naruta]

Just calculate it with X = 10 and Y = 2 and you'll do it

Not really. You´ll have one equation with 2 unknown numbers, which cannot be solves unless you have another equation.

The correct way to solve it is to find the derivative of the function and equal the drervative to 0. And from there on, to obtain the remaining values, like some people did in previous posts.

 

[Strawberry]

Here's the problem. I know you guys are smart enough to figure this out, because I'm not 100% sure on how to do it. Any legitimate attempts will be repped, if I can rep you.

Redo problem 27 in section 4.1 of your textbook (page 178), but find the parabola whose minimum is at (10,2) rather than the point given in the book.

The parabola's equation is y = x^2 + ax + b, where
a =
and b =

put that in vertex form im nota bout to put thought into it but is that all your given?