Consider the following. f(x) = x^2 − 8, (2, −4)
(a) Find the slope of the graph of f at the given point.
I found it to be ((((x+h)^(2)-8)-(x^(2)-8)))/(h)=2x+h. Plugging it in-4.
(b) Find an equation of the tangent line to the graph at the point. (Use the standard coordinate variables x and y.)
(c) Graph the function and the tangent line.
slope of the equation is gieven by the first derivative....in this case derivative of x^2-8 is : 2x
a) 2 (2) = 4 = slope at x = 2
b) tangent line will have this slope y = 2x + b subbing in the point given
-4 = 2(2) +b b = -8 so line equation is y = 2x-8
slope of the equation is gieven by the first derivative....in this case derivative of x^2-8 is : 2x
a) 2 (2) = 4 = slope at x = 2
b) tangent line will have this slope y =4x + b subbing in the point given (typo...I put the wrong slope in there ...calculated '4' but I used '2')
-4 = 4(2) +b b = -12 so line equation is y = 2x-12
Here is the graph:
