Find an equation of the tangent line to the graph of y = g(x) at x = 6 if g(6) = −3 and g'(6) = 5. (Enter your answer as an equation in terms of y and x.)

Answer 1

Use point slope form:

m = 5

(x, y) = (6, -3)

(y + 3) = 5(x - 6)

Answer 2

Find an equation of the tangent line to the graph of y = g(x) at x = 6 if g(6) = −3 and g'(6) = 5. (Enter your answer as an equation in terms of y and x.)

g'(6) = 5 means that the slope of the tangent line to the graph of y = g(x) at (6,-3) is 5. The slope of that line is ALWAYS 5, so for any point on the line, say (x,y), the slope between this point and (6,-3) must be 5 also.

Set the slope from rise over run definition, delta y / delta x = 5

[y-(-3)] / [x - 6] = 5 is such an equation.

Answer 3

The equation of a straight line on a 2-dimensional plane with an x-axis and perpendicular y-axis is of the form:

y = mx + b

where m = the slope of the line and b = the value of the y-intercept of the line.

The y-intercept of a line is the y-value of the (x,y) point on the line for x = 0.

A tangent line is a straight line. The question asks for the equation of the line tangent to the function y = g(x) at x = 6. The problem states that g(6) = -3, which means the point (6,-3) is the point where the tangent line touches g(x). The problem also states that g'(6) = 5, which means the slope of the function g(x) at x = 6 is 5. The tangent line to y = g(x) at this point has the same slope as g(x), so we know m = 5 for the equation of this tangent line.

Inserting y = -3, m = 5 and x = 6 into y = mx + b, provides the value for b:

-3 = 5(6) + b, means b = -33.

So the equation of the tangent line is:

y = 5x - 33