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x^2+y^4=10x+7 @ (1,-2)

Rearrange to isolate y.

y=(-x^2+10x+7)^(1/4)

Take derivative.

y'=(1/4)*((-x^2+10x+7)^(-3/4))*(-2x+10)

y'=(-2x+10)/4*(-x^2+10x+7)^(3/4)

Sub in x=1.

y'=8/2*(16)^(3/4)

y'=2/8

y'=1/4

y'=m=slope=1/4

Sub m and (1,-2) into linear equation (tangent line).

m*(x-x1)=y-y2

(1/4)(x-1)=y+2

Isolate y.

Equation of tangent line: y=(1/4)x-(9/4).

The third question is unclear.

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Q: If x2 plus y4 equals 10x plus 7 then what is the slope of the tangent line at 1 -2 and what is the equation of that tangent line- and at what point are the lines tangent to the original equation?
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