A quadratic function is defined by f(x) = 3x^2 + 4x - 2. A linear function is defined by g(x) = mx-5. What values of the slope o
f the line would make it a tangent to the parabola?
1 answer:
You need to find points where the line g(x) intercepts the quadratic function f(x) in one and only one point.
Then 3x^2 + 4x -2 = mx - 5
solve 3x^2 + 4x - mx -2 + 5 = 0
3x^2 + (4 - m)x + 3 = 0
In order to there be only one solution (one intersection point) the radicand of the quadratic formula must be 0 =>
b^2 - 4ac = (4 - m)^2 - 4(3)(3) = 0
(4 - m)^2 = 24
4 - m = +/- √(24)
m = 4 +/- √(24) = 4 +/- 2√(6)
Then m, the slope of the line, may be 4 + 2√6 and 4 - 2√6
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