SOLUTION: i want to ask how to solve this problems . i want u to explain me the whole question n then give a solution to it. my ques is show that my=x^2-4(x-1) meets the curve y=x^2-3x+2

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: i want to ask how to solve this problems . i want u to explain me the whole question n then give a solution to it. my ques is show that my=x^2-4(x-1) meets the curve y=x^2-3x+2      Log On


   

Question 101124: i want to ask how to solve this problems . i want u to explain me the whole question n then give a solution to it. my ques is
show that my=x^2-4(x-1) meets the curve y=x^2-3x+2 at two distinct points for all non zero values of m.i would be very grateful to u if u solve this ques for me.this ques is nt from text book.






Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
show that my=x^2-4(x-1) meets the curve y=x^2-3x+2 at two distinct points
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y = x^2-4x+4
y = x^2-3x+2
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Substitute to get:
x^2-4x+4 = x^2-3x+2
x= 2
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Substitute to solve for y:
y = 2^2-4*2+4
y = 0
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The curves meet at the point (2,0)
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Cheers,
Stan H.