My maths teacher provided me with a formula. Although i dont think he followed through with the rest of the process. How do i find the equation of the line tangent to x^2, at point (2,4)?
(use dx and dy pls)
2007-05-17 19:05:28 · 3 answers · asked by Idyllic 3 in Science & Mathematics ➔ Mathematics
lol well first you find the derivative (dy/dx) which is 2x. then you plug in the x to find the slope at that point, 4. then just use y=mx+b. 4=4*2+b, b=-4. so the equation is y=4x-4. lol just one question why didn't you ask your teacher?
2007-05-17 19:11:45 · answer #1 · answered by Anonymous · 0⤊ 0⤋
OK, take the derivative of the function. In this case,
y = x^2, so dy/dx = 2x. This is the equation for the line. When x=2, the slope of the line is 4--not because 2^2 =4, but because 2 x 2 =4 (if the original equation had been y=x^3, then dy/dx = 3x^2 and the slope would be 12 (3*2^2) and not 8 (2^3)).
2007-05-17 19:12:25 · answer #2 · answered by Mark S, JPAA 7 · 0⤊ 0⤋
y = f (x) = x²
dy/dx = f `(x) = 2x
At x = 2:-
dy/dx = f `(2) = 4 = gradient , m
Tangent passes thro` (2,4):-
y - b = m.(x - a)
y - 4 = 4.(x - 2)
y = 4x - 4 is equation of tangent at (2,4)
2007-05-17 19:42:15 · answer #3 · answered by Como 7 · 0⤊ 0⤋