H e m P
I'm really stuck on these, if someone could show me the methods for working them out, that would be awesome.
A is the point (-2,1) and B is the point (6,k)
a) show that AB^2 = k^2-2k+65
Given also that AB = 10
b) Find the possible values of K Ok, for a), remember that the distance between two points (a,b) and (x,y) is
distance= ((a-x)^2 + (b-y)^2)^(1/2)
so for the square of the distance, we forget the squareroot on the distance formula and we get
AB^2 = (-2-6)^2 + (1-k)^2
and I'll let you expand it out.
H e m P
Some stuff I've never done before:
c) Find the coordinates of the turning point of the curve with the equation y = 3-5x-2x^2That's all for now. ;_;
Weeellll... how you solve this one really depends on what you know or should know. I'll assume that you don't know how to take a derivative, tell me if you do.
So, start by working out the x intercepts of y = 3-5x-2x^2, which are those x's such that
3-5x-2x^2=0.
Using the quadratic formula, factoring or whatever,
x= -3 or x= 1/2.
Now, parabola's, which are the curves that quadratic functions make, are symmetrical beasts and so, the turning point is half way between the x intercepts and this tells us its at
x= (-3+1/2)/2 = -5/4.
Plug -5/4 into your formula and you'll get the y value and you should be done.
This can also be done by completing the square, and in general that's how you should do it. However, I knew there were nice x intercepts to work with. Anyway,
y = 3-5x-2x^2
= -2(x+5/4)^2 + 49/8
and you can read off the turning point from this formula, its (-5/4, 49/8 ). In gereral, if after completing the square you get
y=a(x-b)^2 +c
then the turning point is at (b,c).
Hope that helps... (and for my own personal reference, don't put an 8 next to a bracket unless you mean it
xp )
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