Find the equation of the tangent line to y = sqrt (x) at x = 9?

Answer 1

Hi,

The equation y = √x is the same as y = x^(½).

Its derivative would be f'(x) = ½x^(-1/2). This is the same as:
..1
------ = f'(x)
√(x)

If I substitute 9 for x in the first derivative, it will give the tangent's slope.

..1
------ = 1/3, so 1/3 is the slope of the tangent.
√(9)

When x = 9, the original equation equals y = √(9) or 3, so the point (9,3) is the point on the graph that the tangent line goes through.

The tangent line's equation in point-slope form is y - y1 = m(x - x1). Filling in our slope of 1/3 and the point (9,3), this becomes:

y - 3 = 1/3(x - 9)
y - 3 = 1/3x - 3
y = 1/3x

This is the equation of the tangent line in slope intercept form for the point of the graph when x = 9.

I hope that helps!! :-)

Answer 2

Take the first derivative of y, then plug in the value x=9; this will give you the slope (m) of the tangent line.

You already know a point on that line (9,3) from the original equation.

Finally, use Point-Slope to find the equation of the line.

y - y1 = m(x - x1)