Show that line y = 2mx - 1 intersects with the curve y = mx^2 - x + m for all non-zero values of m?

7 Answers

  • 1 decade ago
    Best Answer

    Line 1: y = 2mx- 1

    Curve 2: y = mx²- x+ m

    What happens when both L1 and C2 intersect each other? Their (x,y)-values are the same:

    y = y

    2m*x- 1 = m*x²- *x+ m <--- (*x) is the value of x at the intersection point

    I want to know all about m, so let's try out some algebra to split m apart from the rest.

    m*x²- *x+ m = 2m*x- 1

    m*x²- 2m*x+ m = *x- 1

    m(*x²- 2*x+ 1) = *x- 1

    m(*x- 1)² = *x- 1

    m = (*x- 1)/(*x- 1)²

    Nothing was told about *x not being valid at some specific point, so at first *x is valid whatever values it assumes. But notice that m isn't valid when *x equals 1, because doing so the denominator of m equals zero. Let's see what happens when *x approaches 1:

    **limit of m when *x approaches 1 from the right = lim 1/(*x- 1) = 1/0 = +infinite

    **limit of m when *x approaches 1 from the left = lim 1/(*x- 1) = -1/0 = -infinite

    So the bigger the absolute value we choose for m, the more likely the curves are going to intercept each other only around *x = 1. For every value of m, both L1 and C2 are going to intersect each other at more than one point-- except at m = 0 (see details below).

    Since there are no objections against m assuming specific values, ***both curves intersect each other no matter what's the value for "m"***. Even when m equals 0:

    Line 1: y = -1

    Curve 2: y = -x

    Where do both intersect each other? At *x = 1.

  • 1 decade ago

    In order for the line and curve to intersect, there must be a point or point where:

    2mx-1 = mx^2 - x + m


    mx^2 - (2m+1)x +(m+1)=0

    Quadratic formula:

    x = ((2m+1)+/-sqrt(4m^2+4m+1-4m^2-4m)))/2m


    x=1 or x = (m+1)/m

    for all m not equal to zero.

  • Anonymous
    3 years ago

    The function is y = f(x) mean we are in 2d set up. If P and Q are 2 factors on the curve The tangent at P is the restricting posititon of secant PQ as Q has a tendency to P alongside THE CURVE. So first, the question that a secant shifted // to itself does not get up. 2d, secant is a line so we are able to no longer communicate approximately its "length" or a tanget at P to exist, the function would desire to have spinoff defined at P. Geometrically the curve would desire to be mushy. X-axis is in reality tangent to y = x^3 on the muse. The curves like circle and different (mushy) conic sectios supply an effect that the curve would desire to lie completely on a similar edge of the tangent. in spite of the undeniable fact that it no longer so! i think this suffices, Madhukar Sir.

  • 1 decade ago

    2m is the slope of the line

    -1 is the y-intercept

    *The line will always pass the point (0,-1)










    BAck to (0,-1)




    if m=0

    y = mx^2 - x + m will not be a curve anymore but a line...

  • How do you think about the answers? You can sign in to vote the answer.
  • Anonymous
    1 decade ago




    mx^2 +(2m+1)x-m-1=0

    using the cuadratic formula

    we find that:

    x must be either

    -(2m+1) + sqr ( (2m+1)^2-4m(-m-1) )




    -(2m+1) - sqr ( (2m+1)^2-4m(-m-1) )




    there are solutions if the radical (whatever is inside the square root) is not cero:

    (2m+1)^2-4m(-m-1) =4m^2+4m+1 +4m^2+4m

    =8m^2 +8m +1,

    now, this is another quadratic equation, or parabola.

    to check if this equation is ever negative, there are several things you could do. One of them is to find its roots:

    [-8 + sqr(32)]/2

    or [-8 - sqr(32)]/2

    with the help of a calculator (or however you want), you see that

    both are negative numbers, so

    for all m>=0 , 8m^2 +8m +1>=1, in particular positive, therefore

    the original problem will always have 2 solutions.

    please let me know if it is ok, or if somehting is not clear

    Source(s): h
  • Anonymous
    1 decade ago

    Set the two y's equal and solve for x using the quadratic formula.

    The results are: x=1 and x=(m+1)/m m cannot be zero. Q.E.D.

  • 1 decade ago

    Yeah!.... Good luck with that one!

Still have questions? Get your answers by asking now.