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GCSE Maths qs, aargh, help please!!

Can someone explain how you would find the minumum point of the curve:

y=x squared - 8x + 23

:confused:

as it's on a GCSE paper, don't tell me the calculus way, that will confuse me! Also how do you solve equations like:

3/x+3 - 4/x-3 = 5x/x squared - 9

its the A* GCSE stuff i would love to be able to know how to do by monday! :woo:

please try and explain in simple forms, maths isn't my strong point :smile:

thanks x

Reply 1

KingPin

Well on the first point, im very suprised this is at GCSE level, as it is usually AS material- The only way I can think of doing it without using differentiation is graphically
secondly, you need to rearranged the seconded equation so that all the x's are on one side- also note that x^2-9=(x+3)(x-3), which will be the denominator of the other fraction after cross-multiplication

Reply 2

Hollijones

top one you can complete the square to find minimum and maximum points.

Reply 3

Tilly87

Hollijones

top one you can complete the square to find minimum and maximum points.

yeh i did that and got (x-4)^2 + 7, but then i got stuck after that :confused:

Reply 4

Hollijones

so your point is (4,7) you'd done all the hard work lol.

Reply 5

Tilly87

ahah clever, i just need 2 be able to do the other one now :P

Reply 6

Hollijones

i'd try and help but i dont really understand how you've written it. Sorry!

Reply 7

130ss

Draw it. or complete the square, to find the range!!

Reply 8

maxfire

I love it when people can't do maths, and I can.
Makes me feel like a hero.
then I remember,
it's not that cool.

But basically, with a quadratic, you complete the square, if you know how to do that...
then if you have
(x-4)^2 + 7 = y
the minimum point would be 4,7.
because you make the brackets = 0.
so x = 4
and then if x=4, that means y=7

so there you have it.
You're very own minimum point of a curve :'D