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Eyad

  • 3 years ago

Find The equation of the tangent to the curve y=x^2+x^3 at the point (1,0) .

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  1. ash2326
    • 3 years ago
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    @Eyad We have to find the equation of the tangent to the curve \(y=x^2+x^3\) at the point (1, 0) First find \(\frac{dy}{dx}\) at the point (1, 0) this will be the slope of the tangent at (1, 0) and then you can find the equation using this \[\frac{(y-y1)}{(x-x1)}=m\] where \[(x1,y1)=(1, 0) \] and \[\large m= ( \frac{dy}{dx})_{(1,0)}\] Can you try now?

  2. Eyad
    • 3 years ago
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    so its gonna be \[\frac{y-0}{x-1}=0\] ?

  3. Eyad
    • 3 years ago
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    @ash2326 .

  4. ash2326
    • 3 years ago
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    \[\frac{dy}{dx}=3x^2+2x\] put x=1 and y=0 \[\frac{dy}{dx}=3+2=5\]

  5. Eyad
    • 3 years ago
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    Oh Sry i think i need some sleep xD, U did great :D Although its \[2x-3x^2 ...\] where the slope is -1 TY ^_^

  6. ash2326
    • 3 years ago
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    You really are sleepy:P you wrote y=x^2+x^3 instead of y=x^2-x^3

  7. Eyad
    • 3 years ago
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    OH LOL ,Its y=x^2-x^3 .. Night :D

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