Is this math question doable?
This was a question on my math test today. Basically, we were given this function:
f(x) = x^2 + 3x + 2 (so a quadratic or a parabola)
We were told the interval of this function was from -3 to a. We were told to find what a is, assuming that the instantaneous rate of change is -1. How do I go about to doing this? I got it wrong, but I honestly want to know how this question could have been done... Its been bothering me all day.
Thanks for any help!
EDIT: This is for my grade 12 advanced functions class, and my teacher specificed on the test we can't use anything that is taught in Calculus.
October 2, 2013
34 Comments • Newest first
[quote=DarkkBladez]8th grade is algebra 1, if you were in geo in 8th then u were ahead lol[/quote]
oops, you're right hahaha
http://www.wolframalpha.com/input/?i=y+%3D+x^2+%2B+3x+%2B+2+%26+y+%3D+-x+-1
Holy grail of this question. First plot shows a slope of -1 very clearly, and the intersection point.
[quote=wyngstorm]@nsdarknessx: oh it's average rate of change? That makes this doable without calculus. So, the average rate of change is literally the slope of the line connecting the two points. So, in this case, it would be (f(a) - f(-3))/(a - -3)... or a^2 + 3a + 2 - 2/(a + 3), which simplifies to (a^2 + 3a)/(a+3). Then, since ave rate of change = -1, we set this = to -1
a^2 + 3a = a + 3
a^2 + 2a - 3 = 0
Then factor and solve.
(a+3)(a-1)=0
so a = -3 and 1
Since -3 was the original point, a=1 is your answer
Also @alexwee: this is average rate of change, so it doesn't require integration (that's average value of a function). Your first formula was right [/quote]
oh ya.. i mixed the F(x) around with the f(a) and a with x
[quote=wyngstorm]@nsdarknessx: oh it's average rate of change? That makes this doable without calculus. So, the average rate of change is literally the slope of the line connecting the two points. So, in this case, it would be (f(a) - f(-3))/(a - -3)... or a^2 + 3a + 2 - 2/(a + 3), which simplifies to (a^2 + 3a)/(a+3). Then, since ave rate of change = -1, we set this = to -1
a^2 + 3a = a + 3
a^2 + 2a - 3 = 0
Then factor and solve.
(a+3)(a-1)=0
so a = -3 and 1
Since -3 was the original point, a=1 is your answer
Also @alexwee: this is average rate of change, so it doesn't require integration (that's average value of a function). Your first formula was right [/quote]
The way I see the question is a slope of -1 coming off the point x = -3 (instantaneous slope), which connects again to the parabola at a second point, in this case a = -1.
copied from email to TS:
y1(x) = x^2 + 3x + 2
y1(-3) = (9)+ (-9) + 2 = 2, aka (-3 , 2), starting point.
We then know we need a line to connect this point to ( a , b), and that it has a slope of -1. y2(x) = -x + b, since they have to connect at (-3 , 2) we can say
(2) = -(-3) + b;
b = -1;
y2(x) = -x -1
We also know y2(a) = y1(a) is what we need. y1 = y2
x^2 + 3x + 2 = -x -1
0 = x^2 +4x +3
0 = (x + 3)(x + 1)
x = -3, x = -1
x = -3 is obvious since that's what we started with, and x = -1 is our value for a.
[quote=wyngstorm]Well if you have a graphing calculator, graph x^2 + 3x + 2, Find the value at -3 and the value at 1, then find the slope of the line connecting the two points. That's your average rate of change between the two.[/quote]
I just did that and the slope is positive 1.:/ So close.:/
[quote=wyngstorm]He used the wrong formula for calc. Actually, I'm not entirely sure what formula he used? he was setting the first derivative equal 1, which just says that the instantaneous rate of change at a = 1, nothing to do with what you were asking about average rate of change.[/quote]
Is there any method for double check that a=1?
[quote=wyngstorm]@nsdarknessx: oh it's average rate of change? That makes this doable without calculus. So, the average rate of change is literally the slope of the line connecting the two points. So, in this case, it would be (f(a) - f(-3))/(a - -3)... or a^2 + 3a + 2 - 2/(a + 3), which simplifies to (a^2 + 3a)/(a+3). Then, since ave rate of change = -1, we set this = to -1
a^2 + 3a = a + 3
a^2 + 2a - 3 = 0
Then factor and solve.
(a+3)(a-1)=0
so a = -3 and 1
Since -3 was the original point, a=1 is your answer[/quote]
The person before you just did it with calc and got -1.:/
[quote=nsdarknessx]What did you get when you solved it usin algerbra?[/quote]
hrm... i got a answer but it wasn't right after i put it back in...
EDIT: apparently my formula is wrong... so i used calc and found F'(-3)= -3
so to have an avge value of -1 the next value greater than x=-3 has to be 1
so 1=F'(x)
or 1=2x+3
-2=2x
x=-1
or a=-1
[quote=alexwee]@nsdarknessx
ya i think i do but I didn't practice calc for like a year so it might be wrong haha
use the formula of average rate of change or V=change of f(x) over change in x or (F(x)-F(a))/(x-a)
so since we are given x and the average rate of change we'd get this : -1=(F(-3)-f(a))/(3-a)
so input x=-3 and x=a into the F(x) and we'd get: -1=(2-(a^2 +3a+2))/(3-a)
now just use algebra to solve a [/quote]
What did you get when you solved it usin algerbra?
@nsdarknessx
ya i think i do but I didn't practice calc for like a year so it might be wrong haha
use the formula of average rate of change or V=change of f(x) over change in x or (F(x)-F(a))/(x-a)
so since we are given x and the average rate of change we'd get this : -1=(F(-3)-f(a))/(3-a)
so input x=-3 and x=a into the F(x) and we'd get: -1=(2-(a^2 +3a+2))/(3-a)
now just use algebra to solve a
[quote=GoldenStar]If you're thirteen you're nowhere close to doing this type of math unless you skipped 3-4 grades of math.
You should be in geometry/algebra 2 if you live in the US. If that is the case, you'll take
geo-> alg 2/trig -> precalc -> calc[/quote]
8th grade is algebra 1, if you were in geo in 8th then u were ahead lol
[quote=scotthong56]its a parabola, dude, this was 7th grade math.....[/quote]
That's what I thought originally but then read the next line and was like well that escalated quickly.
[quote=nsdarknessx]I just remembered, the question was asking for average rate of change.
So basically, from [-3,a], the average rate of change is -1. You just gotta find a.
Any ideas on how to find a?[/quote]
you punch it in the face and pass out. am bout to sleep
[quote=alexwee]hrm im a little confused at the context of the question
by instantaneous rate of change at -1... is it -1 at a?[/quote]
I just remembered, the question was asking for average rate of change.
So basically, from [-3,a], the average rate of change is -1. You just gotta find a.
Any ideas on how to find a?
hrm im a little confused at the context of the question
by instantaneous rate of change at -1... is it -1 at a?
[quote=scotthong56]its a parabola, dude, this was 7th grade math.....[/quote]
ahahahaha, that made me laugh man.
[quote=LowWillpower]Okay, after a bit of thinking I'm pretty sure I get the question. I at first did not understand why an interval was given, now I get it.
All they want is a line with the slope of -1 to connect from x = -3 to some point a (the instantaneous rate of change between x= -3 and x=a should be -1). So to solve I found that at -3 our equation was equal to 2, from there I knew we had a line of slope -1 with the point <-3,2>, which gave the equation y = -x -1, equating -x - 1 = x^2 + 3x + 2, you can solve for x and get the result of x = -3, x = -1. Obviously x = -3 is where you started, so if x = -1 is your ending point of your interval, the instantaneous change within that interval is the slope which we used to make it, i.e. -1.
Your calculus knowledge just made you think they were asking for something else.[/quote]
wait, how exactly did you solve for x there?
I know it may sound stupid but sometimes this teacher gives back our tests to make one correction, and I wanna make damn sure I get this question corrected.
Okay, after a bit of thinking I'm pretty sure I get the question. I at first did not understand why an interval was given, now I get it.
All they want is a line with the slope of -1 to connect from x = -3 to some point a (the instantaneous rate of change between x= -3 and x=a should be -1). So to solve I found that at -3 our equation was equal to 2, from there I knew we had a line of slope -1 with the point <-3,2>, which gave the equation y = -x -1, equating -x - 1 = x^2 + 3x + 2, you can solve for x and get the result of x = -3, x = -1. Obviously x = -3 is where you started, so if x = -1 is your ending point of your interval, the instantaneous change within that interval is the slope which we used to make it, i.e. -1.
Your calculus knowledge just made you think they were asking for something else.
well if you take the derivative of x^2+3x+2 you get 2x+3. Idk what the question is really asking, if the question is to find the area it covers, just integrate x^2+3x+2 with (-3 and a) so it will be like x^3/3 + 3/2x^2 + 2x and use the B-A law so it will be a^3 + 3/2a^2 + 2a - ( (2^3)( (3/2)(2^2) + (2*2)), then just simplify and get the answer. BAM
you need swag to do this
[quote=nsdarknessx]weird.... maybe we're in the same class.
If you did a similar question, do you wanna share what you did?[/quote]
I had it on a test so I don't remember, sorry.
[quote=ghfusd]@GoldenStar: Well my maths class is quite advanced and we just finished trig.[/quote]
then 1-2 years ^_^
@GoldenStar: Well my maths class is quite advanced and we just finished trig.
Sooooo anyone got the answer yet?
[quote=ghfusd]I'm 13 and what is this?[/quote]
If you're thirteen you're nowhere close to doing this type of math unless you skipped 3-4 grades of math.
You should be in geometry/algebra 2 if you live in the US. If that is the case, you'll take
geo-> alg 2/trig -> precalc -> calc
I'm 13 and what is this?
How did you do Calculus before Advanced Functions?
[quote=Hatchet]Dude what the hell, I live in Brampton too... (I'm serious)[/quote]
weird.... maybe we're in the same class.
If you did a similar question, do you wanna share what you did?
[quote=nsdarknessx]I live in brampton, how about you?
Perhaps our teachers got the answer from the same test bank if you aren't in my class aha?[/quote]
Dude what the hell, I live in Brampton too... (I'm serious)
I'd do that math question so hard
no calculus? challenge accepted.
[quote=Hatchet]This question is really similar to one I did. Are you in my class? o.o[/quote]
I live in brampton, how about you?
Perhaps our teachers got the answer from the same test bank if you aren't in my class aha?
[quote=nsdarknessx]Well the class is called Advanced Functions and the techniques used in Advanced Functions are not the same as the ones used in Calculus, so he told us to stick to techniques taught in advanced functions.
But if you want to use Calculus go ahead. I just want to see what the answer is, and how you got it.[/quote]
This question is really similar to one I did. Are you in my class? o.o
[quote=SpiritBag]Your teacher gave you a calculus problem and said you can't use anything that is taught in Calculus? Ok..[/quote]
Well the class is called Advanced Functions and the techniques used in Advanced Functions are not the same as the ones used in Calculus, so he told us to stick to techniques taught in advanced functions.
But if you want to use Calculus go ahead. I just want to see what the answer is, and how you got it.
Your teacher gave you a calculus problem and said you can't use anything that is taught in Calculus? Ok..