nyansema nyansema
  • 23-08-2017
  • Mathematics
contestada

the equation of the tangent to the curve x^2 = 4y at the point on the curve where x=-2 is?

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kettboy kettboy
  • 23-08-2017
[tex]x^2 = 4y \Rightarrow y= \frac{x^2}{4} [/tex]
[tex] \frac{dy}{dx} = \frac{2x}{4} = \frac{x}{2} [/tex]

(x = -2) [tex] \frac{dy}{dx} = \frac{x}{2} = \frac{-2}{2} = -1[/tex]
(x = -2) [tex]y= \frac{x^2}{4}= \frac{(-2)^2}{4} = 1[/tex]

[tex]y-y_1=m(x-x_1)[/tex]
[tex]y-1 = -1(x--2) [/tex]
[tex]y-1 = -1(x+2)[/tex]
[tex]y-1 = -x-2[/tex]
[tex]y=-x-1 = -(x+1)[/tex]

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Аноним Аноним
  • 23-08-2017
 i think the answer is y=-x-1=-(x+1)
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