Question 1143177: Show algebraically that the line y=kx+5 intersects the parabola y=x^2+2 twice for all values of k. Found 2 solutions by ikleyn, MathLover1:Answer by ikleyn(52754) (Show Source):
The intersection points are those and only those points of the coordinate plane satisfying equation
x^2 + 2 = kx + 5, (1) or
x^2 - kx - 3 = 0. (2)
The discriminant of the equation
d = " b^2 - 4ac " = k^2 - 4*(-3) = k^2 + 12
is positive for any real value of "k".
It means that equation (2), and hence equation (1), has two distinct real roots for all values of k.
Hence, the line y=kx+5 intersects the parabola y=x^2+2 twice for all values of k.
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Geometrically, the point (x,y) = (0,5) lies on the given straight line and "inside" of the parabola.
Therefore, it is clear that the given line (the class of lines) intersects the parabola at two different points.
You can put this solution on YOUR website! Show algebraically that the line intersects the parabola twice for all values of .
----------------- ........solve for using quadratic formula
-> since a=1,b=-k and c=-3, we have
For your problem the important part of this expression is the discriminant,
and if is positive there are
since in your case , if there will be two solutions
........solve for ....as you can see could be number (positive or negative) and will be greater than which is your proof
so, is element of
check few solutions:
let's
graph:
as you can see, the line intersecting parabola twice
let's
graph:
as you can see, the line intersecting parabola twice
let's
graph:
as you can see, the line intersecting parabola twice
