Given a function y = g(x) such that g(1) = 4 and g'(1) = -1, find the equation for the tangent line to the graph of y = g(x) at the point where x = 1 in point-slope form.

Answer 1

since g(1) = 4, you want the line through (1,4) with slope -1
y-4 = -1(x-1)

Answer 2

you need 2 things to find the equation of the tangent line at a point to y: the point (x,y) and the slope at x

point: (1, 4) which is given from g(1) = 4

slope of tangent line @ x = 1: m = -1, which is given by g'(1) = -1
*remember, the slope of the tangent line is equal to the first derivative

now we have the point and the slope of the tangent line at that point, we can plug it into the point-slope equation

equation:
y - y1 = m(x - x1), where y1 = y coordinate of the point, m = slope of the tangent line aka the first derivative at the point, x1 = x coordinate of the point
plug in the values we found above: y - 4 = -1(x - 1)

*take note of which form of the equation the question wants you to find, because it can sometimes be standard or slope-intercept form, BUT always solve for point-slope form first and then translate into other forms