y=√(25-x^2)
y' = -x/√(25-x^2)
y'(3) = -3/4
so, now you have a point and a slope. The line is thus
y-4 = -3/4 (x-3)
find an equation of the line tangent to y= sqrt(25-x^2) at the point (3,4)
i got y= x/8 + 29/8
this is wrong. but im not sure why. can you explain thank you
1 answer