Answer:
and 
Step-by-step explanation:
The equation of curve is

We need to find the equation of the tangent line to the curve at the point (-3, 1).
Differentiate with respect to x.
![2[2(x^2+y^2)\frac{d}{dx}(x^2+y^2)]=25(2x-2y\frac{dy}{dx})](https://tex.z-dn.net/?f=2%5B2%28x%5E2%2By%5E2%29%5Cfrac%7Bd%7D%7Bdx%7D%28x%5E2%2By%5E2%29%5D%3D25%282x-2y%5Cfrac%7Bdy%7D%7Bdx%7D%29)

The point of tangency is (-3,1). It means the slope of tangent is
.
Substitute x=-3 and y=1 in the above equation.





Divide both sides by 130.

If a line passes through a points
with slope m, then the point slope form of the line is

The slope of tangent line is
and it passes through the point (-3,1). So, the equation of tangent is


Add 1 on both sides.


Therefore,
and
.
Answer:
x=-1 is the answers for the question
Step-by-step explanation:
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Answer: The Answer is A
Step-by-step explanation:
i think it's right
Answer:
Step-by-step explanation:
g(x)= -x-1 or g(x) = x+1
if g(x) = -x-1
-x-1=2x+4
-x=2x+5
-x/2=x+5
-x/2 -x=5
-x/2-2x/2=5
-x-2x=5
-3x=5
x=5/-3
if g(x) = x+1
x+1=2x+4
x=2x+3
x/2=x+3
x/2-x=3
x/2-2x/2=3
x-2x=3
-x=3
x=3
When x=0 y=50 when y=0 x=+\-5. -x^2 makes the parabola open downward y(x)=2x^2+50;y(3)=2*9+50=32