Aquadratic 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 of the line would make it a tangent to the parabola?

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

no solution!