the equation of the tangent line to the graph of f(x) at (5, 52) is y=mx+b for m=25 and b=? f(x) = 3x^2-5x+2
m = f'(x) sooo y = (6(5) -5)(x-5)+52
thanks, how did you get that?
well, they already gave the slope so the derivative was just a throw in
but given a point P(x,y) and a slope of m; the equation can be determined by: y = m(x-Px)+Py
ahh, I see what I was doing wrong.
where Py = the b part...
you can get there quicker by: y = mx -mPx + Py maybe :)
yeah, that worked. thanks again
youre welcome :)
how would you get the intercept for this one then? f(x)=\[\sqrt{6-x}\] m=-.25
the slope is given by the derivative at x=2
crap, not slope. I meant intercept.
the intercept is rather vague; do you mean the x intercept of the tangent line to the curve at x=2?
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