a. find an equation for the secant line through the points where x has the given values.

Question

b. find a equation for the line tangent to the curve when x has the first value.
y=9square root(x); x=16, x=25

Answer 1

y = 9√x
y(16) = 36
y(25) = 45
So, the secant has slope (45-36)/(25-16) = 1
So, now use the point-slope form to get the equation for a line with slope 1 through (16,36)

For the tangent line, the slope

y' = 4.5/√x is 9/8 at x=16

Again, use the point-slope form to get the tangent line