help out :) calculus

Question

Use the implicit differentiation method to find the equation of the line that is tangent to the curve c ^ (- (x ^ 2 + y)) = x + y + 1 at the point P(1, 1).

Answer 1

-x^2-y = x+y+1

x^2-x -1= 2y

y= (.5)(x^2 +x)-.5

day/dx= x+1/2= slope

at x=1 dy/dx = 1+1/2 = 3/2, but (1,1) is not on the parabola

there's a problem with the problem interpreted that way

plot the graph and point. there are 2 lines fitting the requirements

with slope = dy/dx = (.5x^2 +.5x -.5 -1)/(x-1)= 3/2

x^2 +x-2-3= 3x-3

x^2 -2x =0= x(x-2), x=0, 2, y= -.5, -2 (0, -.5), (2,-2)

line through (0,-.5) & (1,1) is y-1=6(x-1)

line through (2,-2), (1,1) is y-1= -3(x-1) in point slope forms

Answer 2

-2x - y' = 1 + y'; y' = - (2x + 1)/2;

Y = 1 - (2x + 1)/2(X - 1);

y = 1 - (2x + 1)/2(x - 1), 2y = 2 - 2x² + 2x - x + 1; 2y = x + 3 - 2x²;

2y = - x² - x - 1; - x² - x - 1 = x +3 - 2x²; .x² - 2x - 4 = 0; x = 1 ± √5

x = 1 +√5; 2y = - (1 + 2√5 +5 +1 + √5 + 1); y = - 4 - 3√5/2

x = 1 - √5; 2y = - (1 - 2√5 + 5 +1 - √5 + 1); y = 4 - 3√5/2

This is 2 tangent points and because now easy to get equations of 2 tangent lines.

;