DLS
Find the equation to the tangent to the curve x^2+3y3=0 which is parallel to the line y=4x5



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

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

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my attempt^

amistre64
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why x=1/2 ?

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oops 1/2

amistre64
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other than that, your process is good

DLS
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12y48x+13=0 is what im getting :O

amistre64
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y = 5/4
y+5/4 = 4(x1/2)
this is a line equation ... so if form does not matter this would suffice

amistre64
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otherwise, algrbrate it to your hearts content

DLS
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isnt y 11/12?

amistre64
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.5^2 + 3(.5)  3
.25 + 1.5  3
1.75  3 = 1.25 = 5/4

amistre64
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since x = 1/2 and NOT 1/2 you have to reevaluate the y you found

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

amistre64
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yes, which is what i just did

DLS
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okay! thanks! got it

amistre64
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\[\frac14+\frac32\frac31\]
\[\frac14+\frac64\frac{12}4\]
\[\frac{712}4\]

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

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

amistre64
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ohh
then your derivative might need some looking at:
x^2+3y3=0
2x +3y' = 0
y' = 2x/3 = 4
x = 12/2 = 6

amistre64
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(6)^2+3y3=0
33+3y=0, when y = 11

amistre64
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y+11 = 4(x+6) would be the tangent line

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(6,11)

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CORRECT NOW! :D

amistre64
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yay! ;)