Equation of the tangent line to the curve at given point? help

Find an equation of the tangent line to the curve at the given point.

1.) y equals 2x to the power 3 - 5x, (-1,3)


2.) y equals sqrt of x, ( 1,1)


Plz show work.had to spell out some symbols since there not on my Android key pad.

To find the slope, take the derivative of the function

slope = y' = 6x^2 - 5
at point (-1,3), slope = y'(-1) = 6(-1)^2 - 5 = 1

plug this information into point-slope form of a line
(y-3) = (1)(x-(-1))
y = x + 1 +3
y = x+4

For the second equation:
slope = y'(1) = 1/(2sqrtx) = 1/2

(y-1) = 1/2 (x-1)
y = 1/2x + 1/2