How do you find an equation of the tangent line, in slope-intercept form, to the curve #y=(2x+3)^(1/2)# at the point x=3? b.) Find the equation of the normal line to the above curve at x=3.?

To find the equation of the tangent line to the curve y=(2x+3)^(1/2) at the point x=3, we need to find the derivative of the function and evaluate it at x=3. The derivative of y=(2x+3)^(1/2) is dy/dx = (1/2)(2x+3)^(-1/2). Evaluating this at x=3, we get dy/dx = (1/2)(2(3)+3)^(-1/2) = (1/2)(9)^(-1/2) = 1/6.

The slope of the tangent line is equal to the derivative at the given point, so the slope of the tangent line is 1/6.

Using the point-slope form of a line, y - y1 = m(x - x1), where (x1, y1) is the given point (3, (2(3)+3)^(1/2)), and m is the slope (1/6), we can substitute these values into the equation to find the equation of the tangent line.

Simplifying the equation, we get y - (2(3)+3)^(1/2) = (1/6)(x - 3).

To find the equation of the normal line to the curve at x=3, we need to find the negative reciprocal of the slope of the tangent line. The negative reciprocal of 1/6 is -6/1 = -6.

Using the point-slope form of a line again, y - y1 = m(x - x1), where (x1, y1) is the given point (3, (2(3)+3)^(1/2)), and m is the slope (-6), we can substitute these values into the equation to find the equation of the normal line.

Simplifying the equation, we get y - (2(3)+3)^(1/2) = -6(x - 3).