If we have the curve y = sqrt(x), find the y value and the slope of the curve when x = 36. y = 6 Correct: Your answer is correct. slope = 1/12 Correct: Your answer is correct. Hence, find the equation of the tangent line to the curve at x = 36, writing your answer in the form y = mx + c. What are the values of m and c?m = c =

Answer:Equation of tangent of curve at x = 36:     1)[tex]y = \frac{x}{12} + 3[/tex] 2[tex]y = \frac{-x}{12} - 3[/tex]     Step-by-step explanation: We are given the following information:[tex]y = \sqrt{x}[/tex]Value of curve when x =  36:[tex]y = \sqrt{36} = \pm 6[/tex]Thus, [tex]y = \pm6[/tex], when x = 6.Slope of curve, m = [tex]\frac{dy}{dx} =\frac{d(\sqrt{x})}{dx}=\frac{1}{2\sqrt{x}}[/tex]At x = 36, slope of curve =[tex]\frac{1}{2\times \sqrt{36}}\\\\m=\frac{1}{12},\frac{-1}{12}[/tex]Equation of tangent of curve at x = 36:[tex](y-y_1) = m(x-x_1)[/tex] [tex]= (y-(\pm 6)) = (\pm\frac{1}{12} )(x - 36)[/tex]Thus, equation of tangents are:1)[tex](y-6) = \frac{1}{12}(x-36)\\12(y-6) = x-36\\y = \frac{x}{12} + 3[/tex]Comparing to [tex]y = mx + c[/tex], we get [tex]m = \frac{1}{12}[/tex] and [tex]c =3[/tex]2)[tex](y+6) = \frac{-1}{12}(x-36)\\12(y+6) = -x+36\\y = \frac{-x}{12} - 3[/tex]Comparing to [tex]y = mx + c[/tex], we get [tex]m = \frac{-1}{12}[/tex] and [tex]c =-3[/tex]