Find the equation of the tangent line to the graph of

Question

 f (x) = √(x^2 + 3) at the point (-1, 2)? 

Answer 1

Slope - intercept form of the line equation is y = mx + b, where m is slope and b is y - intercept.

The curve f(x) =√(x² + 3) and the point is (- 1, 2).

y = √(x² + 3)

Differentiating  the curve with respect to x.

y' = {1/[2√(x² + 3)]} * [x² + 3]'

y' = (2x)/[2√(x² + 3)]

y' = x/√(x² + 3).

At (- 1, 2), y' = (- 1) / √[(- 1)² + 3]

y' = - 1/√[1 + 3]

y' = - 1/√4

y' = - 1/2.

This is the slope(m) of the tangent line.

m = - 1/2.

Now, the line equation is y = (- 1/2)x + b.

Find the y - intercept by substituting any point in the line equation say (x, y) = (- 1, 2).

2 = (- 1/2)(- 1) + b

b = 2 - 1/2

b = (4 - 1)/2

⇒ b = 3/2.

The tangent line equation is y = (- 1/2)x + (3/2).