Given the parabola y = x & # 178; + 4 and the straight line y = x + 10, find: (1) the intersection of them; (2) the tangent equation of the parabola at the intersection. Solve it by derivative

(1)
x^2+4=x+10
x^2-x-6=0
(x-3)(x+2)=0
X = 3 or - 2
Y = 13 or 8
The intersection is (3,13) (- 2,8)
(2)f'(x)=2x
f'(3)=6,f'(-2)=-4
The intercept of two tangents are
13-6*3=-5 ; 8-(-2)*(-4)=0
The two tangent equations are
y=6x-5
y=-4x

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