The Student Room Group

axes of symmetry

in my book it says evaluate all axes of symmetry in these graphs.

for a I completed the square
for b I just realised that X=2 is the only solution therefore is the only axes of symmetry however

question 3 and 4 I don't know how to work this out.

do I complete the square? I'm not to sure I've tried quite a lot

IMG_20181118_212526-compressed.jpg.jpeg
Original post by utv
in my book it says evaluate all axes of symmetry in these graphs.

for a I completed the square
for b I just realised that X=2 is the only solution therefore is the only axes of symmetry however

question 3 and 4 I don't know how to work this out.

do I complete the square? I'm not to sure I've tried quite a lot


For b) it should be x=2x=-2.

For (c) notice that you can rewrite it as x=(y3)24x = (y-3)^2 - 4 which is a completed square form of a horizontal quadratic as opposed to your usual \cup or \cap shape one like in part (a). So can you see what the axis of symmetry there is?

For (d) you may notice that x=4x=4 or x=3x=-3 are the axes of symmetry, but it's actually the line that is in the middle of the two.
Reply 2
thanks for your help but for c wouldn't you have to make it =0 to use complete the square?
Original post by utv
thanks for your help but for c wouldn't you have to make it =0 to use complete the square?


It already IS in completed square form once you move the 4 across. The square is completed in the variable yy here, not xx.

As I said, this is a sideways parabola which means it has equation x=f(y) x = f(y) as opposed to the usual vertical parabolas of the form y=f(x)y = f(x).
Reply 4
oh ok great thanks it makes sense now. Ty

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