Question

What's the equation of the tangent to
`y = x^2` at
`x = 2`?

Answer 1

y=4x-4.

Explanation:

1. the required equation of the tangent line has the form:

y=kx+i, where k=f'(x₀), i - interception;

2. if x₀=2, then y₀=2²=4;

3. f'(x)=2x, then f'(x₀)=2*2=4;

4. the required equation with unknown 'i' is y=4x+i;

it is possible to calculate unknown 'i' using coordinates (x₀;y₀):

5. 4=4*2+i, ⇒ i=-4;

6. finally, y=4x-4.

P.S. the suggested solution is not the shortest one.