# maths

1)y=x²-3x+2
2)y=8x-16/x-3

A)Sketch both on axes
B)Calculate coordinate of points of intersection

I have drawn both on axes. I am having trouble with B). I know you have to do y=y.

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3. 👁 47
1. Well, the first one is a parabola of course.
I complete the square to graph

x^2 -3 x = y - 2
x^2 - 3 x + 9/4 = y+ 1/4
(x-3/2)^2 = y+ 1/4
so vertex at (3/2, -1/4)
now zeros
x = 3/2 +/- (1/2)sqrt (9-8)
= 1 or 2 at y = 0

However the second one if it is really

y = 8 x -(16/x) - 3

must be hard to sketch because it is undefined at the origin.
It goes through (1, -11)
It goes through (2,5)
so it crosses the parabola somewhere between x = 1 and x = 2
look for that point somewhere around 1.25
also look for a point at larger x
and check for a point at negative x
I would do it numerically but you can try setting them equal
x^2 - 3 x + 2 = 6 x - 16/x -3
x^3 - 3 x^2 +2 x = 6 x^2 -16 - 3x
x^3 - 9 x^2 + 5 x +18 = 0
or
x^3 - 11 x^2 + 5 x = -16
try x around 1 to 2 and look for -16
x = 1 --> -5
x = 1.2 --> -8.1
x = 1.3 ---> -9.89
x = 1.4 ---> -11.8
x = 1.5 ---> -13.9
x = 1.6 ---> -16 Oh, oh score
Now you could call (x-1.6) a factor and divide to get a quadratic or just search the same way or other roots.

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posted by Damon
2. I don't know if your second equation is y = (8x-16)/(x-3) or 8x - (16/x) -3 or 8x - 16/(x-3)
Without parentheses, it is ambiguous.
Whatever the second equation is, if you have drawn the graphs, just see where they intersect. Or, as you mentioned, solve the equation y1 = y2, where the y's are the two different formulas.

y2 = (8x-16)/(x-3) = 8(x-2)/(x-3),
Then use also
y1 = (x-2)(x-1)
Then (x-2)(x-1)= 8(x-2)/(x-3) at the intersections.
x-1 = 8/(x-3)
x^2 -4x -5 = 0
(x-5)(x+1) = 0
x = 5 is one solution. At that intersection point, y = 25 -15 +2 = 12
I'll leave the other for you to solve.

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posted by drwls
3. Remember x = 2, both sides 0

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posted by Damon

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