Submitted by Tim on Mon, 08/22/2016 - 18:55

Hi all! This is what I've done for part vii)

y=x²+2kx where 2≥x≥2 and -2>k>2

When k>0 let k=a, when k<0 let k=-a

When k>0, y=x²+2ax

This can be rewritten as:

y=(x+a)²-(a²) or y=x(x+2a)

Therefore the parabola passes through the origin and -2a with minimum at (-a,-(a²)). Hence, the smallest value = -(a²)

f(2) = 4+4a

f(-2) = 4-4a

Therefore 4+4a is the greatest value.

When k<0, y=x²-2ax

This can be rewritten as:

y=(x-a)²-(a²) or y=x(x-2a)

Therefore the parabola passes through the origin and 2a with minimum at (a,-(a²)). Hence, the smallest value = -(a²) (as before)

f(2) = 4-4a

f(-2) = 4+4a

Again, as before, 4+4a is the greatest value.

As -2>k>2 I can say:

-(a²)2 then a>2

so as a --> infinity,

-(a²) --> -(infinity)

4+4a --> infinity

Is this also a suitable answer?

I don't have answers for these questions so the best method I can think of is posting what I've done onto the forum and hoping everyone else is getting the same thing. xD

If something isn't quite right then please say! Thanks!

## My post got cut a bit...

My post got cut a bit. I'm not sure what happened there.

"As -2>k>2 I can say:

-(a²)2 then a>2

so as a --> infinity,

-(a²) --> -(infinity)

4+4a --> infinity

Is this also a suitable answer?"

Is meant to say:

"As -2>k>2 I can say:

-(a²)2 then a>2

so as a --> infinity,

-(a²) --> -(infinity)

4+4a --> infinity

Is this also a suitable answer?"

## Intervals

I've only scanned your work, and I am speaking very loosely, but I'm presuming you are puzzled because your answer is disappearing off to infinity, which doesn't feel right.

I'd say, have a close look at the intervals in the question.

This all presumes - amongst other things - I've read a) your post and b) the question correctly.

Anyway, I hope it helps.

## k

a cannot tend to infinity! It is in the range [-2,2]. Your answers for the greatest and least values should really be in terms of k and not your variable.

You don't actually have to consider k negative and positive separately when completing the square, and if you know a bit about modulus functions you can write expressions for the greatest and least values which work for all k.

(None of the answers in this assignment are infinty or negative infinity...)

## Oops

Right I'm on it, my bad. Thanks for the help! ^^

## Least answers

Are the solutions for the minimum values meant to be in terms of $k$ and $x$ or am I missing a trick?

Currently I have the coordinates of the minimum which are $(-kx, -(kx)^2)$ but I'm unsure if I need to somehow factor $x$ out.

## Clarification

My current solution for the least value is $k^2x(x-2)$.

The reason I ask is just because I find it conceptually somewhat difficult to picture what the solution would be for a known value of $k$.

## And I Found the Mistake

I was - for some inexplicable reason - of the opinion that $x^2 + 2kx = (x+kx)^2 - (kx)^2)$.

Not quite sure how I managed to do that.

## Least values

The least value should be in terms of $k$. If $-2 2$ then the vertex is not in the given range of x, so the minimum will be at one end of the range or the other (so at x=2 or x=-2).