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.
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.
3 answers
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.
If your second equation is
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.
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.
If your second equation is
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.
Remember x = 2, both sides 0