LINE tangent to the graph?

Question

Determine an equation of the line tangent to the graph of the following equation at hte indicated point
f(x) = (2x-4)^2; (4,16)

Please i will appreciate if you show me how you got to the answer

Answer 1

The quick answer is to take the derivative of f(x) ...

f'(x) = 2 (2x - 4) 2 = 4 (2x - 4)

... and evaluate it at x = 4
to get a slope for the tangent line of
f'(4) = 16. [Ask if you need to know
why / how to do this.]

Now you want the equation of a line through (4, 16) with slope 16. That would be

(y - 16) = 16 (x - 4)

or

y = 16 x - 48

Check: it has slope 16 and passes through (4, 16).

Answer 2

first of all notice that (4,16) is a point on the curve y=(2x-4)^2.

The slope of the tangent line to the curve at x=4 is equal to the derivative f'(x)=4(2x-4) at x=4.
f'(4)=16

This is the m in the lines equation passing through (4,16):
y-16=m(x-4)
Substitute m=f'(4)=16 we have
y-16=16(x-4)
y-16=16x-64
16x-y=64-16

16x-y=48 or y=16x-48

Answer 3

wow I tryed and I came up with the same aswer but not this way thanks for asking this question.