*At what point is the tangent to the curve `x+y^2=1` parallel to the line `x+2y=0` ?*

(1) To find the slope of the tangent line to the curve we find its first derivative. We can use implicit differentiation.

`d/(dx)[x+y^2]=d/(dx)1`

`1+2yy'=0`

`y'=-1/(2y)` . Thus if we know a point on the...

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*At what point is the tangent to the curve `x+y^2=1` parallel to the line `x+2y=0` ?*

(1) To find the slope of the tangent line to the curve we find its first derivative. We can use implicit differentiation.

`d/(dx)[x+y^2]=d/(dx)1`

`1+2yy'=0`

`y'=-1/(2y)` . Thus if we know a point on the graph, the slope of the tangent line is given by `-1/2` of the reciprocal of the y-coordinate.

(2) The slope of the line `x+2y=0` is `-1/2` .

(3) If the slope of the tangent line is parallel to the slope of the given line, then they have the same slopes.

Thus `-1/(2y)=-1/2` , so y=1. Substituting for y in the equation of the curve gives x=0.

**So the tangent line to the given curve at (0,1) is parallel to the given line.**

(** We could have written the curve as two functions; `y=+-sqrt(1-x)` and differentiated each function. Using implicit differentiation avoids a number of potential pitfalls from this method **)

The curve is a parabola with vertex at (1,0) opening to the left. A quick visual check affirms the reasonableness of the answer.