The discriminant of a quadratic equation, ax^{2} + bx + c = 0 is D = b^{2} – 4ac. If D>0, the quadratic equation has two distinct roots; if D<0, then the equation has no real roots; and, if D=0, the we have two equal roots. Let’s apply it in the following problem.

**What is the equation of the line tangent to f(x)=x**^{2} at the point (2,4)?

^{2}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, x^{2} = 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 x^{2} – mx – k = 0 is a quadratic with two equal roots, that is, its discriminant D = b^{2} – 4ac = 0 or D = m^{2} – 4(1)(-k) = 0. This implies that k = -m^{2}/4. We can then write x^{2} – mx – k = 0 as x^{2} – mx + m^{2}/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 = -m^{2}/4: k = -4. The equation of the line is therefore y=4x-4.

###### Solution using the idea of factors of perfect square trinomial

We can still solve the problem even if you cannot remember the discriminant of quadratic equation. Because there is only one point of tangency, we only have one solution to the equation x^{2} = mx + k. Thus, x^{2} – mx – k = 0 is a perfect square. Remember that trick about completing the square where we halved the coefficient of the middle term and then square it. We can then write x^{2} – mx – k = 0 as x^{2} – mx – (-m^{2}/4) or x^{2} – mx + m^{2}/4 = 0. From here we proceed as the above solution.

###### Solution using derivative

Typically, we use the derivative to solve this type of problem. To calculate for the slope of the tangent we get the derivative of f(x) which is f'(x)=2x. Now, if x=2, then 2x=4 =m which is the slope of the tangent line. From here we substitute it to the general equation of the line which is y=mx+k using the point (2,4) to calculate for k: 4=4(2)+k which yields k=-4. So, the equation of the line is y=4x-4.

Here’s another version of the problem: Point of tangency – no calculus, please.

the whole concept is explained in a very simple way.

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