# `f(x) = 1/sqrt(x - 1), x + 2y + 7 = 0` Find an equation of the line that is tangent to the graph of f and parallel to the given line.

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Since the tangent line to function f(x) = 1 / sqrt(x -1) has to be parallel to the line x + 2y + 7 = 0, both needs to have the same slope.

Equation of the straight line can be rewritten in slope-intercept form to : y = -x/2 - 7/2. Hence the desired slope of our line is -1/2.

And derivative of f(x) will provide the slope of the line tangent to f(x)

d/dx(f(x)) = -1/2 * (x-1) ^ (-3/2)

setting it equal to -1/2

-1/2 * (x-1) ^ (-3/2) = -1/2 or x = 2 provides one of the conditions

Corresponding y coordinate can be found from the f(x) = 1/sqrt(x - 1) equation, y = 1/sqrt(2 - 1) = 1

The slope of the desired line is equal to -1/2, and we now have a point on the desired line (2, 1), as well the point of tangency

Using the point-slope, equation to desired line can be obtained as follows. Given slope m = -1/2 and a point on the desired line,(x0, y0), the equation of the line is given by

(y - y0) = m(x - x0)

y - 1 = -1/2(x - 2 )

y = -1/2x + 2

2y + x - 4 = 0

The given equation of the line x+2y+7=0 can be transformed to found the slope of the line.

2y=-x-7

y=(-1/2)x-7/2

Therefore the slope(m) of the line is -1/2.

The slope of the line that is tangent to the graph is the derivative of the graph.

f'(x)=(-1/2)(x-1)^(-3/2)

Since the tangent line to the graph and the given line are parallel , therefore the slope of the both lines will be equal.

(-1/2)(x-1)^(-3/2)= -1/2

(x-1)^(-3/2)=1

(x-1)^(3/2)=1

x-1=1

x=2

Therefore y coordinate can be found by substituting x in the function

y=1

The equation of the line that is tangent to the graph and parallel to the given line can be found by the point slope form of the line.

y-y_1=m(x-x_1)

y-1=(-1/2)(x-2)

2y-2=-x+2

**2y+x-4=0**