Question

How to find the inverse of y=x^2-x by completing the square?

Answer 1

well can you complete the square in x...?

Answer 2

I'm not sure.. I thought that we had to swap the x and y then solve for y then?

Answer 3

well the way the question reads... you need to complete the square 1st

Answer 4

\[y\quad =\quad { x }^{ 2 }-x\]

Answer 5

so the 1st take is to halve the coefficient of x, which is -1

then square the answer

so its

\[(\frac{-1}{2})^2\]

what would the value be..?

Answer 6

i'm still confused on how to go on :/

Answer 7

is it (x-1/2)^2=0?

Answer 8

ok... so if you have a quadratic

\[x^2 + bx \]

to complete the square you need to add

\[(\frac{b}{2})^2\]
to complete the square

so in your question

b = -1

so you need to add

\[(\frac{-1}{2})^2\] to get the perfect square...

does that make sense

Answer 9

yep.. i get that

Answer 10

ok to balance the equation and keep it in its original form you need to also subtract that value

so you have

\[y = (x^2 -x + \frac{1}{4}) - \frac{1}{4}\]

is that ok...?

Answer 11

ahh, i see.. yep yep

Answer 12

ok.. so factoring the brackets you get

\[y = (x - \frac{1}{2})^2 - \frac{1}{4}\]

is that ok...?

Answer 13

yeah :)

Answer 14

ok... now to find the inverse... swap x and y.. so you have

\[x = (y - \frac{1}{2})^2 - \frac{1}{2}\]

so the task to get the inverse is use the above equation and make x the subject.

Answer 15

oops it should read...

to get the inverse... make y the subject

Answer 16

and we solve for y now, yeah?

Answer 17

thats it... sorry for the typo

Answer 18

it's all good

Answer 19

hope it helps you find the inverse

Answer 20

yep! I got it! Thank you!!!