How do you find the equation of the line tangent to the graph of #y = sqrtx# at (16,4)?
1 Answer
May 15, 2018
Explanation:
#"to obtain the equation we require the slope m and a "#
#"point on the tangent"#
#"the point "(x_1,y_1)=(16,4)#
#m_(color(red)"tangent")=dy/dx" at x = 16"#
#y=sqrtx=x^(1/2)#
#rArrdy/dx=1/2x^(-1/2)=1/(2x^(1/2))=1/(2sqrtx)#
#x=16tody/dx=1/(2xx4)=1/8#
#rArry-4=1/8(x-16)#
#rArry=1/8x+2larrcolor(red)"equation of tangent"#
