VCE Methods Question Thread!

[knightrider]

How would you do this question?

A rectangular area, 14m by 16m is to be divided into 4 equal garden beds.Between each section there is a x-metre wide path.

find a formula for the total area of the path?

[Zues]

ive forgetten this,

can someone show me the steps when graphing this f(x) = square root (16-x^2)

e.g. like x intercepts, y int.

i get an x value of 4, but answer has -4

[Zues]

how to find range of this?

[Zues]

with domain is it always left to right
with range is it always bottom to top? even if the square root function for example has an end point at (-2,0) (where it begins) and finishes at -2. is the range (-2,0) ? (forget round/square brackets)

[Zues]

final one haha.

why is the range ...,square root 3) , because it passes through this should it have a square bracket?

or is this only applicable to end points?

[Sunshine98]

ive forgetten this,

can someone show me the steps when graphing this f(x) = square root (16-x^2)

e.g. like x intercepts, y int.

i get an x value of 4, but answer has -4

f(x) = square root (16-x^2) ,  can be re written as  square root (-x^2+16) ,
 if u remove your negative sign your left with the expression -(x^2-16) ,
factorise and you get -(x-4)*(x+4) ( all under the square root sign , of course)
 so ur now left with f(x) = square root  -(x-4)*(x+4) which can also be written as f(x) = square root  (x+4)*(x-4) (Ive  rewritten it because we cant have a negative under square root)
solve for x , and as u found out x is positive 4 , and as Ive shown here , can be negative 4.
Hope that answers ur question.
Soz for not using that fancy equation writer , I don't know how to use it

[Zues]

f(x) = square root (16-x^2) ,  can be re written as  square root (-x^2+16) ,
 if u remove your negative sign your left with the expression -(x^2-16) ,
factorise and you get -(x-4)*(x+4) ( all under the square root sign , of course)
 so ur now left with f(x) = square root  -(x-4)*(x+4) which can also be written as f(x) = square root  (x+4)*(x-4) (Ive  rewritten it because we cant have a negative under square root)
solve for x , and as u found out x is positive 4 , and as Ive shown here , can be negative 4.
Hope that answers ur question.
Soz for not using that fancy equation writer , I don't know how to use it

i can see that, it can be positive 4 or negative 4, but why does it start at negative 4?

[keltingmeith]

i can see that, it can be positive 4 or negative 4, but why does it start at negative 4?

What do you mean "why does it start at negative 4?"?

Also, on your many range questions, the short answer is to draw the graph and look at what y-values you have. For the more specifics, I think you're just unsure on drawing the graphs given the domain, which is something you should practice so that you can get the range questions down.

[Zues]

alright can you explain this one then?

drawing this on the cas, it starts at (0,2) hence the range is [2, infinity] why is it (2, infinity) ie. B

[theshunpo]

alright can you explain this one then?

drawing this on the cas, it starts at (0,2) hence the range is [2, infinity] why is it (2, infinity) ie. B

The domain is R+, therefore 2 isn't included as you cannot input 0

[cosine]

alright can you explain this one then?

drawing this on the cas, it starts at (0,2) hence the range is [2, infinity] why is it (2, infinity) ie. B

Well when you draw it on your CAS, you can see that the graph stops, and does not go through the y-intercept at (0,2), and because it does not go through the point, it is not included.

[cosine]

how to find range of this?

I keep getting this as my range. [6,infinity)
The answer is [5, infinity) and they already give you the domain as (-infinity, 3]
Someone please show me were i am going wrong?
Thanks

[IndefatigableLover]

I keep getting this as my range. [6,infinity)
The answer is [5, infinity) and they already give you the domain as (-infinity, 3]
Someone please show me were i am going wrong?
Thanks

Take note that the given domain may not always yield the 'lowest' or 'highest' points for the range... sometimes when you graph it out, it's easier to see. For instance, the turning point is located at x=2 where the y value is 5 (which happens to be the lowest point and within the given domain).

[cosine]

Take note that the given domain may not always yield the 'lowest' or 'highest' points for the range... sometimes when you graph it out, it's easier to see. For instance, the turning point is located at x=2 where the y value is 5 (which happens to be the lowest point and within the given domain).

OH so you are saying that although y=6 is the first y value, it doesnt matter, so to find out the range we have to consider the LOWEST y value to the HIGHEST?

[theshunpo]

OH so you are saying that although y=6 is the first y value, it doesnt matter, so to find out the range we have to consider the LOWEST y value to the HIGHEST?

Exactly