The equation of the tangent line to the curve y = √(2x - 1) that is parallel to the line x - 3y = 16 can be determined by first finding the slope of the given line, which is 1/3. Next, the derivative of the curve, y' = 1/(√(2x - 1)), must be set equal to 1/3 to find the corresponding x-value. Substituting this x-value back into the original curve equation yields the point of tangency, which can then be used in the point-slope form y - y0 = m(x - x0) to derive the final equation of the tangent line.
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The slope-intercept form is often written as y = mx + b.DespicableMe said:Homework Statement
What is the equation of the tangent line to the curve y = Sqrt( 2x - 1) where the tangent line is parallel to the line x - 3y = 16
The Attempt at a Solution
My teacher briefly mentioned to find the derivative of the given, and then plug it into the slope-y intercept formula? I got confused from there because he used y-y=m(x-x)