Garis singgung

What is the equation of the line tangent to f(x)=x² at the point (2,4)?

Solution using the idea of discriminant

The function f(x) and the line y=mx+k are equal, that is, x² = mx + k at the point of tangency. It implies that the solution or value of x of this equation is unique and in fact is equal to the x coordinate of the point at the point of tangency.  It follows that x² – mx – k = 0 is a quadratic with two equal roots, that is, its discriminant D = b² – 4ac = 0 or D = m² – 4(1)(-k) = 0. This implies that k = -m²/4. We can then write x² – mx – k = 0 as x² – mx + m²/4 = 0 which is equivalent to                  (x – m/2)(x – m/2)= 0. Using the principle that if ab=0, the a=0 or b= 0, we can say that x-m/2 = 0. This implies that x=m/2. We know from the given that x = 2 so m = 4. We can then solve for k in the equation k = -m²/4: k = -4. The equation of the line is therefore y=4x-4.
( http://math4teaching.com/2014/04/17/application-discriminant/)