Hi can someone help with this maths question

Okay... I suppose you want to find the actual quadratic.

I can't follow, from the info given, what do you mean by "(x-1)(x-3)"?
Also, can you use calculus? (P.S. You really don't need it, but some people prefer calculus, so I'd ask just in case)

No I cant use calculus, the topic is quadratic functions

Yes from your example that would be (h,k)

Good, now from the info given from the question, you know "h" and "k" of the quadratic in completed square form.
What you don't know is "a"... what else have you not used?

We are practically done.

The (0,-5) maybe

Yeah, can you complete the problem now? You should be able to do this.

Just in case you don't know how (0, -5) comes into play...

I did this
(x-2)²+3
But when I expand it, the y intercept is 7 not -5

Ah, common error here.
(x-2)²+3 is not a sentence - it's missing a verb (always be mindful writing maths is nothing different than writing english, it's just muddled with symbols).
In this case, it's the equal sign.

Well, what does (x-2)²+3 equal to? Is it really 0? Reread #4 carefully.

I'm sorry I'm very confused

It's fine.

At this point, at best we can say the quadratic is in the form y=a(x-2)^2+3 (reread #4, there's an "a" in front to be found).

Recall from #6 that we only know h=2 and k=3. We don't know what "a" is.
You (unintentionally) assumed it's 1, which isn't always the case.
Also remember (0,-5) was not used? It's not meant to be used as double-checking, it actually gives information (on what "a" is).

P.S. Problems are often structured such that you need every single piece of information given with no extras. That is, no more, and no less.

Ok how would u do that because I get this part
(X-2)²+3
X²-4x+7
But what can a be that would give -5
Sorry for taking so long to understand but I have no clue how to work out the a part

Again, at best we know it's y=a(x-2)²+3, not y=(x-2)²+3!
Do actually copy down the whole thing, don't skip anything (i.e. don't forget the "y=" and the "a").
Then proceed with using (0,-5) by...


P.S. A good solution structure, at least from my habit of doing maths, would look a bit like this...