Find two values of k, such that the line y = kx + 2 is tangent to the curve y = x^2 + 4x + 3
There will be intersection when x^2 + 4x + 3 = kx + 2. Our goal is to find the values of k which would only give one solution to this quadratic equation, which would make the lines 'tangent' to each other. First we rearrange the equation to get it in a familiar form: x^2 + (4-k)x + 1 = 0 To have one solution, the discriminant (b^2 - 4ac) must be zero. (4-k)^2 - 411 = 0 16 - 8k + k^2 - 4 = 0 k^2 - 8k +12 = 0 Factorising: (k-6)(k-2) = 0 So k = 6 and k = 2 are valid solutions to this problem.
AN
Related Maths A Level answers
All answers ▸Consider the functions f and g where f (x) = 3x − 5 and g (x) = x − 2 . (a) Find the inverse function, f^−1 . (b) Given that g^−1(x) = x + 2 , find (g^−1 o f )(x) . (c) Given also that (f^−1 o g)(x) = (x + 3)/3 , solve (f^−1 o g)(x) = (g^−1 o f)(x)
Find the normal to the curve y = x^2 at x = 5.
Express x^2-7x+2 in the form (x-p)^2+q where p and q are rational. Hence or otherwise find the minimum value of x^2-7x+2
Write down the values of (1) loga(a) and (2) loga(a^3) [(1) log base a, of a (2) log base a of (a^3)]