A quadratic function is defined by

f(x)= 3x^2+4x-2. A linear function is defined by g(x)= mx-5. what values(s) of the slope of the line would make it a tangent to the parabola

To find the values of the slope (m) that would make the line a tangent to the parabola, we need to determine when the linear function intersects the quadratic function at exactly one point.

For the line to be tangent to the parabola, it means that the line should touch the parabola at one point only, which implies that the discriminant (b^2 - 4ac) of the quadratic equation should be equal to 0.

Given the quadratic function:
f(x) = 3x^2 + 4x - 2

Comparing it with the standard form of a quadratic equation (ax^2 + bx + c), we have:
a = 3
b = 4
c = -2

To find the value(s) of slope (m), we substitute the values from the quadratic equation into the linear equation g(x) = mx - 5.

g(x) = mx - 5

Substituting the values of a, b, and c:
g(x) = mx - 5 = 3x^2 + 4x - 2

Now, to find the discriminant, we set the quadratic equation equal to the line equation:
3x^2 + 4x - 2 = mx - 5

Rearranging the equation to set it equal to 0:
3x^2 + (4 - m)x - (2 + 5) = 0
3x^2 + (4 - m)x - 7 = 0

The discriminant is given by b^2 - 4ac:
(4 - m)^2 - 4(3)(-7) = 0

Simplifying the equation:
(16 - 8m + m^2) + 84 = 0
m^2 - 8m + 100 = 0

Now, we can solve this quadratic equation for m by factoring or using the quadratic formula.

Using the quadratic formula:
m = (-b ± √(b^2 - 4ac)) / (2a)
m = (-(-8) ± √((-8)^2 - 4(1)(100))) / (2(1))
m = (8 ± √(64 - 400)) / 2
m = (8 ± √(-336)) / 2
m = (8 ± √(336)i) / 2
m = 4 ± 2√(21)i

Therefore, the values of the slope (m) that would make the line a tangent to the parabola are m = 4 + 2√(21)i and m = 4 - 2√(21)i.