Find the range of values of k for which the line y=kx+9=0 does not intersect the curve y=x²-2x?
Find the range of values of k for which the line y=kx+9=0 does not intersect the curve y=x²-2x
Is there anything wrong with this question? I cannot seem to solve it.
I found it very weird too, why is an equation, y=kx+9=0 given instead of an expression. This question appeared in my Math exam for the first question!
- Anonymous8 years agoBest Answer
yes, there's something wrong with this question. y = kx + 9 = 0 just does not make sense. If you meant y = kx + 9, then there is no k value such that y = kx + 9 does not intersect y = x^2 - 2x
- icemanLv 78 years ago
y = x^2 - 2x => the parabola, vertex at(1 , -1) , the domain is all the real numbers, but the range is limited by y value since the global minimum is at y = -1, the the range is:
y ≥ -1
Now, for the line y = kx + 9 not to intersect with the parabola y must be less tha -1 or
kx + 9 < -1
kx < -10
k < -10/x
- anordtugLv 68 years ago
y=x^2-2x is a kind of parable coming from + values of y when x is -. It goes throgh 0 when x=0.
Then it is going down until x=1 and start growing.(dy/dx=0 at x=1)
it is negative from x=0 until x=2, minimum point (1,-1)
Therefore kx+9 must pass under (1,-1) and becase this line cross y-axis at +9 this is a hopeless problem to solve.
- 8 years ago
any value of 'k' will do the job i.e, there is no value of 'k' for the line to intersect the given curve