Finding the Vertex of a Translated Curve

  • Thread starter Thread starter Trail_Builder
  • Start date Start date
  • Tags Tags
    Confusion Vector
Join the discussion
Registration is free. Start your own thread to ask a follow-up.
8 replies · 5K views
Trail_Builder
Messages
148
Reaction score
0
hi, I'm basically a bit confused as to how to solve this vector question, in other words, in what i have to do which is confusing me...

Homework Statement



This is a sketch of the curve with equation y = f(x)
It passes through the origin O.

(the sketch has no value on or anything, its simply a happy parabola then passes through the origin on the left side a bit of the way up, and dips in the negative y values)

The only vertex of the curve is at A (2, -4)

a) Write down the coordinates of the vertex of the curve with the equation (bold are my answers)

i) y = f(x-3) (5, -4)
ii) y = f(x) -5 (2, -9)
iii) y = -f(x) (2, 4)
iv) y = f(2x) (1, -4)

The curve with equation y=x^2 has been translated to give the curve y=f(x)[/]

b) find f(x) in terms of x

Homework Equations



(I don't know if the first half of the question is needed, if it is, say so and ill post it up)

The Attempt at a Solution



not really sure, i know how functions work, I know all about vectors too, but not sure what it means by "translated", and how i would find out the answer.

my instinct says the answer is simply x^2, but somehow I don't think it is, and i certainly don't think that is worthy of 4 marks :S


hope you can help, thnx


i added in the entirity of the question now..
 
Last edited:
Physics news on Phys.org
Translated how? The question as posted doesn't make much sense. Better post the rest.
 
added to the topic the entirity of the question, hope that helps
 
Ok, so y=f(x) is a parabola going through the origin with a vertex at (2,-4) and you want to find its equation given that its a translation of y=x^2? Well, you want to move it two units right and four units down. Having done part a) you should be able to manage that. What's your guess?
 
Trail_Builder said:
hi, I'm basically a bit confused as to how to solve this vector question, in other words, in what i have to do which is confusing me...

Homework Statement



This is a sketch of the curve with equation y = f(x)
It passes through the origin O.

(the sketch has no value on or anything, its simply a happy parabola then passes through the origin on the left side a bit of the way up, and dips in the negative y values)

The only vertex of the curve is at A (2, -4)
That, together with the fact that it is a "happy" (I like that!) parabola with vertex (2, -4) tells you that y= a(x-2)2- 4 for some positive number a.

a) Write down the coordinates of the vertex of the curve with the equation (bold are my answers)

i) y = f(x-3) (5, -4)
ii) y = f(x) -5 (2, -9)
iii) y = -f(x) (2, 4)
iv) y = f(2x) (1, -4)

The curve with equation y=x^2 has been translated to give the curve y=f(x)[/]
Okay, that tells you that the "a" above is 1!

b) find f(x) in terms of x
From the first statement, that this is a parabola opening upward with vertex (2, 4), you know the equation is y= a(x-2)2+ 4. Now you know that a= 1!

Homework Equations



(I don't know if the first half of the question is needed, if it is, say so and ill post it up)

The Attempt at a Solution



not really sure, i know how functions work, I know all about vectors too, but not sure what it means by "translated", and how i would find out the answer.

my instinct says the answer is simply x^2, but somehow I don't think it is, and i certainly don't think that is worthy of 4 marks :S


hope you can help, thnx


i added in the entirity of the question now..
 
o rite, see how the answer is y=f(x-2)+4 now, but one thing i a tad confused on

the a you refer to, that's the stretching and stuff yeh, and just to check, is the reason you came to the conlusion that it has to be 1, because it said y = x^2 is translated. does that term translated refer specificially to it moving not being strechted or whatever?

thnx for the help
 
Yes, in connection with movement (as opposed to translating a language!) means a movement with no stretching or rotating.

By the way, why did you title this "vector question confusion" (and refer to it as a "vector question" in your post) when there are no vectors involved?
 
HallsofIvy said:
Yes, in connection with movement (as opposed to translating a language!) means a movement with no stretching or rotating.

By the way, why did you title this "vector question confusion" (and refer to it as a "vector question" in your post) when there are no vectors involved?


oops haha, yeh i realized what i did


basically, i was stuck writing a vector problem in the topic, but then i figured out what i was doing, then went onto to the next question (this question), and forgot to change the topic title hehe

oops