A calculus problem by Lew Sterling Jr
Calculus
Level
2
f(x) = sqrt{x^{2}+3} at the point (-1, 2). -Find the equation of the tangent line.
x + 2y = -3
x - 2y = 3
-x + 2y = -3
x + 2y = 3
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1 solution
Hello & peace be upon you,
as f(x) = ( x^2 + 3 )^0.5
m = the slope = f'(x) = x / ( x^2 +3 )^0.5 at x = -1,
therefore, m = -1 / ( 1 + 3 )^0.5 = 1/2 = 0.5
use the formula,
y - y1 = m( x- x1) as given the point, (-1,2),
y - 2 = 0.5 ( x +1 )
2y - 3 = x
x + 2y = 3,
thanks and have a good day....