f(4)=1 f ' (x)=-2 Find an equation of the tangent line to f(x)at the point x=4.
to represent a line by algebraic forms, you need to know two things. slope and the point it passes. those informations are in the f(4)=1 f'(x)=-2
I have to find the slope and y intercept so I don't have any of that either
I have no idea where to start I really need help please anyone
y-f(4)=f'(x)(x-4) y-1=-2(x-4) 2 x+y-9=0
so then the 2x+y=0 is the tangent equation
oh sorry I mean 2x+y-9=0 is the tangent equation
is this correct
you understand that f'(a) is a slope of the tangent line at x=a right?
and f'(x)=-2 says f'(4)=-2
also, y intercept is just one of the points that line passes. you can use other points that line pass. f(4)=1 says the tangent line passes (4,1)
so assuming that you understand y=m(x-a)+b, y=(-2)(x-4)+1
or in other form, 2x+y=9
ok so then are yousaying that the slope is -2 and the y intercept is 1
defintion of y intercept is the value of y at x=0
y intercept is of course 9..
ok yup just got that so then the slope is -2
yeah
ok got that then what it I have t estimate f(7) then what so I do
i don't understand what you want to know. what is t?
sorry bad typing I meant how do I find the estimate of f(7)
oh so estimation is something close to the real solution right?
yes
in this case, (assuming you don't know integral calculus) you can't figure out the exact f(7) because you don't know the rule
right it just has to be an estimate
yeah. well you can use your tangent line to make that approximation
ok so how do you do that
you plug 7 into x in your equation of tangent line
actually the whole point of calculus is that if you cut curves very small, it estimates to straight lines
so then is the esimate of f(7)=-5
oh ok
it is called 'linear approximation' and your answer is right
yay ok thanks for all your help
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