Question

calculus help please
find the lines that are (a) tangent and (b) normal to the curve y=sqrt(x) at x=4

Answer 1

y=x^1/2

Answer 2

find derivative

Answer 3

1/2sqrtx
put x=4
1/(2)(2)
slope=1/4

Answer 4

of tangent at x=4

Answer 5

ok so the slope is 1/4 so then the equation of the tangent would be
y-2=(1/4)(x-4) which is really y=(1/4)x +1??

Answer 6

and then since the equation of the normal has an opposite and reciprocal slope would the equation of the normal be
y=-4x+16??

Answer 7

y=2 you are right

Answer 8

for normal line
take - inverse of slope of tangen

Answer 9

ok..so now would this be correct?
EQ tangent = y=(1/4)x +1
EQ normal = y=-4x+16

Answer 10

slope of normall =-4
pass through 4,2
y-2=-4(x-2)
y=-4x+8+2
y=-4x+10

Answer 11

ok that makes perfect sense. i guess i did my quick algebra incorrectly.
thank you so much you have been a great help

Answer 12

welcome