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

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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.

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

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

Rearranging:

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

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

x=((2m+1)+/-1)/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.

• 2m is the slope of the line

-1 is the y-intercept

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

Substitute...

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

0=m+1+m

0=2m+1

-1=2m

-1/2=m

y=2mx-1

y=2(-1/2)x-1

y=-x-1

BAck to (0,-1)

0=-(-1)-1

0=0

But...

if m=0

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

• Anonymous

sure

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

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

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

we find that:

x must be either

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

------------------------------------------------

2

or

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

------------------------------------------------

2

so

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