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# slope of graph at given point

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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.

Mar 23, 2021

#1
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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

Mar 23, 2021

#1
0

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

Guest Mar 23, 2021
#2
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2x-8 isn't correct, I don't know why it's wrong

Guest Mar 23, 2021
#3
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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:

Guest Mar 23, 2021
#4
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Double sorry      (didn't fix the last equation in my answer:

so line equation is y  = 4x-12

Mar 23, 2021