Find the value of the constant for which the line is a tangent to the curve
I can't continue because Im getting a negative value inside the square root, where did I go wrong?
They're not saying that y+2x=k is the equation of the curve, but that it is tangent to the curve.
This means that at a given point, the tangent to the curve will have y+2x=k as an equation.
The tangent to the curve at a point of absciss a is :
Hence, calculate the derivative of and then... think
Thus, is tangent to if and only if has exactly one solution. That means, the quadradic has exactly one solution. That happens precisely when the discriminant vanishes, thus, .
Here is how you do it.. Easy and Simple
Line: y = k-2x
Curve: y = x²-6x+14
k-2x = x²-6x+14
x²-6x+2x+14-k = 0
x²-4x+14-k = 0
a=1, b=-4, c=(14-k)
[I]Use this Equation[/I] ---- b²-4ac
4k = -16+56
4k = 40
k = 40/4
k = 10 <---- Here's your Answer!!!