Calculus determine functions formula from graph

[algray09]

For part a I think you can use the information provided. At first I thought I could use vertex to start finding the formula but it’s not whole number. Could I use 2 of the points to find slope to start?

Part b I think to find formula I could use the two points to find slope. Maybe the formula has (2,1) in it because it’s the vertex.

 

[jonah2.0]

Beer induced suggestion follows.

...
Part b I think to find formula I could use the two points to find slope. Maybe the formula has (2,1) in it because it’s the vertex.

Google the phrase "given three points determine parabola equation".

 

[algray09]

Beer induced suggestion follows.

Google the phrase "given three points determine parabola equation".

Ok so I did and followed the steps for part b I know c=2 but following the steps that google said I got 12a+2b=-1.
Here were my steps that I did
1.a(0)^2+b(0)+c=2, a(2)^2+b(2)+c=1, a(4)^2+b(4)+c=2

a(0)+b(0)+c=2, a(4)+b(2)+c=1, a(16)+b(4)+c=2


2. (4a+2b+c=1) - (0a+0b+c=2) = 4a+2b=-1

3. (16a+4b+c=2) - (0a+0b+c=2) = 16a+4b=0

4. (16a+4b=0) - (4a+2b=-1) = 12a+2b=-1

Checking my work I looked it up and correct answer for part b is g(x)= x^2/4 - x +2. I don’t know what to do because I think the steps I followed is wrong because that doesn’t get me to a or b

 

[Steven G]

(0)-(-1)=-1 is not correct.
You already had two independent equations involving a and b, so why bother getting a 3rd one?
Students tend to forget that a prerequisite for Calculus is algebra and arithmetic!

 

[algray09]

(0)-(-1)=-1 is not correct.
You already had two independent equations involving a and b, so why bother getting a 3rd one?
Students tend to forget that a prerequisite for Calculus is algebra and arithmetic!

For part b there are 3 points the steps on google say to make 3 equations one for each set of points so that’s why I had 3

But I figured it out now part a is f(x)= -2/3x^2 -4x -16/3.
Part b is g(x)= 1/4x^2 -x+2

 

[Steven G]

For part b there are 3 points the steps on google say to make 3 equations one for each set of points so that’s why I had 3

But I figured it out now part a is f(x)= -2/3x^2 -4x -16/3.
Part b is g(x)= 1/4x^2 -x+2

You did have 3 equations--- a(0)^2+b(0)+c=2, a(2)^2+b(2)+c=1, a(4)^2+b(4)+c=2
But you knew that c=2. That leaves two unknowns, namely a and b. That is why you then only needed two equations.

 

[topsquark]

I don't like this question at all. You can make a good guess at what the functions are but you can't tell for sure without more information.

-Dan

 

[Deleted member 4993]

I don't like this question at all. You can make a good guess at what the functions are but you can't tell for sure without more information.

-Dan

I agree. We can "estimate" that g(x)_min =1 at x = 2

 

[Dr.Peterson]

I don't like this question at all. You can make a good guess at what the functions are but you can't tell for sure without more information.

-Dan

Agreed: they should make it explicit that the functions are quadratic. Students should not be taught that guessing is mathematics.

@algray09: As for the two functions specifically, I would use the factored form for f, and vertex form for g, rather than the general form as you are doing.

 

[JeffM]

I fully agree with those who have pointed out that the problem given does not specify that the functions are quadratic. Without that clue, it is just an awful problem. Of course, if it comes from a chapter discussing parabolas specifically or conic sections generally, this is not so horrible a question as it seems superficially. The OP did not seem in any doubt in the very first post that it involved quadratics.

 

[Otis]

None of the comments above occurred to me, but I had noted they hadn't asked for domain statements also (despite providing graphs that don't entirely span the horizontal axis).

One of my pet peeves
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