Question

Find the inverse
y=x^2+5x


x=y^2+5y
x=y(y+5)
Now I'm stuck

Answer 1

You're right about the
\[\large x = y^2 + 5y \] however, a problem occurs... we need a technique called 'completing the square' have you heard of it?

Answer 2

Completing the Square 101
Imagine you have something of the form
\[\Large z^2 + \color{purple}pz = k\]
And for some reason or another, you need to express it in terms of a perfect square trinomial.

The first thing to watch out for is to make sure that the square bit... (this part)\[\Large \color{red}{z^2 }+ \color{purple}pz = k\] is unadorned... meaning, it has

no number next to it... if it does, well, remedy it by dividing the entire expression by that number...

So now, pay attention to the coefficient of the 'unsquared' variable, in this case, \(\large \color{purple}p\). What you need to do is halve it, resulting in \(\Large

\frac{\color{purple}p}{2}\) and then square it, yielding \(\Large \frac{\color{purple}p^2}{4}\).

Next, you add this to both sides of your equation, like so..
\[\Large z^2 + \color{purple}pz+\frac{\color{purple}p^2}4 = k+\frac{\color{purple}p^2}{4}\]

You'll notice that the left side is now a perfect square, in particular, the square of \(\Large z+\frac{\color{purple}p}2\)...
\[\Large \left(z+\frac{\color{purple}p}2\right)^2=k+\frac{\color{purple}p^2}{4}\]

And that's it!
You have successfully 'completed the square'
:)

Answer 3

so I would take
x=y^2+5y
make it
x+5y/2=(y+5/2)^2

Answer 4

Whoops...

Answer 5

No... lol what the coefficient of the y is just 5. What's half of that?

Answer 6

5/2

Answer 7

And the square of that is...

Answer 8

25/4

Answer 9

and THAT's what you add to both sides.

Answer 10

so it would be
x+25/4=(y+5/2)^2

Answer 11

That's correct :)
\[\Large x + \frac{25}4 = \left(y+\frac52\right)^2\]
I reckon it's easier to find the inverse now? :)

Answer 12

Still lost? ^_^

Answer 13

so it would be
\[f ^{-1}(x)=\sqrt{x+\frac{ 25 }{ 4 }}\] minus 5/2

Answer 14

Precisely!
\[\Large f^{-1}(x) = \sqrt{x+\frac{25}4}- \frac52\]

Well done!

Answer 15

ok so I have one more it is \[y=2^{3-x}\]
I am thinking \[\log_{2}3-y \]
Im just not sure if I can subtract the y then subtract x and divide by -1 to get the inverse

Answer 16

Let's star at the very beginning...
\[\Large y = 2^{3-x}\]
first, switch y and x...
\[\Large x = 2^{3-y}\]
Then I suppose you took the log(base 2) of both sides?

Answer 17

\[\Large \log_2(x)=3-y\]

Answer 18

And then just solve for y.

Answer 19

ok I see what you're saying and what I did wrong thank you for all of your help

Answer 20

No problem ^_^