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lyudmila [28]
3 years ago
13

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

Mathematics
2 answers:
aleksandrvk [35]3 years ago
7 0
x^2 = 4y \Rightarrow y= \frac{x^2}{4}
\frac{dy}{dx} = \frac{2x}{4} =  \frac{x}{2}

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

y-y_1=m(x-x_1)
y-1 = -1(x--2)

y-1 = -1(x+2)
y-1 = -x-2
y=-x-1 = -(x+1)

Yuliya22 [10]3 years ago
6 0
 i think the answer is y=-x-1=-(x+1)
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