Note: I am assuming that the equation is `y = x^2 + kx +3` .If it were a cubic function, the given line will never be tangent to it.
For a line to be tangent to a curve represented by a quadratic equation, there must only be one intersection.
Equating the two (y is isolated):
`2x + 1 = x^2 + kx + 3`
`x^2 + kx - 2x + 3 - 1 = 0`
`x^2 + (k-2)x + 2 = 0`
The value(s) of k for which the line is tangent to the curve are precisely those that will result to a single root to this equation. For a quadratic equation to have a single root, its discriminant,D, must be equal to zero.
`D = b^2 - 4ac`
where `b = k-2` ,`a = 1` ,and `c = 2` .
`D = (k-2)^2 - 4(1)(2) = 0`
`(k-2)^2 = 8`
`k-2 = pm sqrt(8)`
Case 1: `k-2 = +2sqrt(2) ` yields `k = 2+2sqrt(2)` .
Case 2: `k-2 = -2sqrt(2)` yields `k = 2 - 2sqrt(2)` .
Hence, for the line to be tangent to the curve, k must either be `2+2sqrt(2)` or `2-2sqrt(2)` .
no, the equation is `y=x^3 + kx + 3` ,