`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.
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.
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.
Since the tangent line to the graph and the given line are parallel , therefore the slope of the both lines will be equal.
Therefore y coordinate can be found by substituting x in the function
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.