Asked by Madi
1. use the definition mtan=(f(x)-f(x))/(x-a) to find the SLOPE of the line tangent to the graph of f at P.
2. Determine an equation of the tangent line at P.
3. Given 1 & 2, how would I plot the graph of f and the tangent line at P if : f(x)=x^2 +4, P(4,20)
2. Determine an equation of the tangent line at P.
3. Given 1 & 2, how would I plot the graph of f and the tangent line at P if : f(x)=x^2 +4, P(4,20)
Answers
Answered by
Reiny
first of all, if you are trying to define the definition of the derivative, it should be
mtan=Lim (f(x+a)-f(x))/(a) , as a --->∞
Furthermore, I will assume that f(x) = x^2 + 4, you only state that at the very end
f(x+a) = (x+a)^2 + 4
then (f(x+a)-f(x))/(a) = (x^2 + 2ax + a^2 + 4) - x^2 - 4)/a)
slope of tangent = lim (x^2 + 2ax + a^2 + 4 - x^2 - 4)/a as a --->∞
= lim (2ax + a^2)/a , as a --->∞
= lim a(2x + a)/a , as a --->∞
= lim 2x + a , as a --->∞
= 2x
now at the point (4,20)
the slope of the tangent is 2(4) or 8
and the tangent equation is
y-20 = 8(x-4)
y = 8x - 12
check with Wolfram:
http://www.wolframalpha.com/input/?i=plot+y+%3D+x%5E2+%2B+4,+y+%3D+8x+-+12
mtan=Lim (f(x+a)-f(x))/(a) , as a --->∞
Furthermore, I will assume that f(x) = x^2 + 4, you only state that at the very end
f(x+a) = (x+a)^2 + 4
then (f(x+a)-f(x))/(a) = (x^2 + 2ax + a^2 + 4) - x^2 - 4)/a)
slope of tangent = lim (x^2 + 2ax + a^2 + 4 - x^2 - 4)/a as a --->∞
= lim (2ax + a^2)/a , as a --->∞
= lim a(2x + a)/a , as a --->∞
= lim 2x + a , as a --->∞
= 2x
now at the point (4,20)
the slope of the tangent is 2(4) or 8
and the tangent equation is
y-20 = 8(x-4)
y = 8x - 12
check with Wolfram:
http://www.wolframalpha.com/input/?i=plot+y+%3D+x%5E2+%2B+4,+y+%3D+8x+-+12
Answered by
scott
a --->∞ should probably be a --->0
Answered by
Reiny
Scott, of course you are right, my silly error.
Fortunately, my calculations match the a --->0 conditions
Fortunately, my calculations match the a --->0 conditions
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