DLS
  • DLS
Find the equation to the tangent to the curve x^2+3y-3=0 which is parallel to the line y=4x-5
Mathematics
jamiebookeater
  • jamiebookeater
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DLS
  • DLS
\[\LARGE \frac{dy}{dx}=2x+3=4\] \[\LARGE x=-\frac{1}{2},y=\frac{11}{12}\]
DLS
  • DLS
\[\LARGE (y-\frac{11}{12})=4(x+\frac{1}{2})\]
DLS
  • DLS
my attempt^

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amistre64
  • amistre64
why x=-1/2 ?
DLS
  • DLS
oops 1/2
amistre64
  • amistre64
other than that, your process is good
DLS
  • DLS
12y-48x+13=0 is what im getting :O
amistre64
  • amistre64
y = -5/4 y+5/4 = 4(x-1/2) this is a line equation ... so if form does not matter this would suffice
amistre64
  • amistre64
otherwise, algrbrate it to your hearts content
DLS
  • DLS
isnt y 11/12?
amistre64
  • amistre64
.5^2 + 3(.5) - 3 .25 + 1.5 - 3 1.75 - 3 = -1.25 = -5/4
amistre64
  • amistre64
since x = 1/2 and NOT -1/2 you have to reevaluate the y you found
DLS
  • DLS
arent we substituting x=1/2 in the equation of curve?
amistre64
  • amistre64
yes, which is what i just did
DLS
  • DLS
okay! thanks! got it
amistre64
  • amistre64
\[\frac14+\frac32-\frac31\] \[\frac14+\frac64-\frac{12}4\] \[\frac{7-12}4\]
DLS
  • DLS
@amistre64 sorry,it was 3Y and not 3X that is how I got 11/12...
DLS
  • DLS
it is x squared so -1/2 and 1/2 both are same that is why i didnt change the y component
amistre64
  • amistre64
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
amistre64
  • amistre64
(6)^2+3y-3=0 33+3y=0, when y = -11
amistre64
  • amistre64
y+11 = 4(x+6) would be the tangent line
DLS
  • DLS
(-6,-11)
DLS
  • DLS
CORRECT NOW! :D
amistre64
  • amistre64
yay! ;)

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