Question

Find the equation to the tangent to the curve x^2+3y-3=0 which is parallel to the line y=4x-5

Answer 1

\[\LARGE \frac{dy}{dx}=2x+3=4\]
\[\LARGE x=-\frac{1}{2},y=\frac{11}{12}\]

Answer 2

\[\LARGE (y-\frac{11}{12})=4(x+\frac{1}{2})\]

Answer 3

my attempt^

Answer 4

why x=-1/2 ?

Answer 5

oops 1/2

Answer 6

other than that, your process is good

Answer 7

12y-48x+13=0 is what im getting :O

Answer 8

y = -5/4

y+5/4 = 4(x-1/2)

this is a line equation ... so if form does not matter this would suffice

Answer 9

otherwise, algrbrate it to your hearts content

Answer 10

isnt y 11/12?

Answer 11

.5^2 + 3(.5) - 3

.25 + 1.5 - 3

1.75 - 3 = -1.25 = -5/4

Answer 12

since x = 1/2 and NOT -1/2 you have to reevaluate the y you found

Answer 13

arent we substituting x=1/2 in the equation of curve?

Answer 14

yes, which is what i just did

Answer 15

okay! thanks! got it

Answer 16

\[\frac14+\frac32-\frac31\]
\[\frac14+\frac64-\frac{12}4\]
\[\frac{7-12}4\]

Answer 17

@amistre64 sorry,it was 3Y and not 3X that is how I got 11/12...

Answer 18

it is x squared so -1/2 and 1/2 both are same that is why i didnt change the y component

Answer 19

ohh

then your derivative might need some looking at:

x^2+3y-3=0

2x +3y' = 0

y' = -2x/3 = 4

x = 12/-2 = -6

Answer 20

(6)^2+3y-3=0

33+3y=0, when y = -11

Answer 21

y+11 = 4(x+6) would be the tangent line

Answer 22

(-6,-11)

Answer 23

CORRECT NOW! :D

Answer 24

yay! ;)