The first equation, which is a straight line, intersects the second equation, which is a parabola, in 3 ways ...
- it misses the curve, no solution
- it cuts the curve in 2 distinct points, 2 solutions
- it touches the curve in 1 point, one solution
It is the last case we want.
To have a single solution the quadratic must factor into two equal factors of the form (qx+b)^2
first equation: y = 2 - 3x
plug into second
(4x-3)(x-2) = 2-3x
4x^2 -8x + 4 = 0
x^2 - 2x + 1 = 0
(x-1)^2 = 0
one solution ..... x = 1
sub back into y = 2-3x
y = 2-3 = -1
the point is (1,-1)
(check my arithmetic)
how do you show that line 3x+y-2=0 is a tangent of the curve y=(4x-3)(x-2) and how to find the point of contact??
1 answer