1) We take the function and derivate it: y′=dxdy=dxd(−2x2−4x)=−4x−4=−4(x+1)
The slope m1 of the tangent line will be: y′(−3,−6)=−4(−3+1)=−4(−2)=8=m1
We need to use the relation to find the slope of the normal line m2: m1m2=−1→m2=−m11=−81
We use the point (-3, -6) to find the whole equation:
y−y0=m(x−x0)→y−(−6)=−81(x−(−3))→y+6=−8x−83→y=−8x−83−6=−8x−851
2) We take the function and derivate it:
y′=dxdy=dxd(−x3+2x2−3)y′=−3x2+4x=x(4−3x)
The slope m1 of the tangent line will be: y′(1,−2)=(1)(4−3⋅1)=1(1)=1=m1
We need to use the relation to find the slope of the normal line m2: m1m2=−1→m2=−m11=−11=−1
We use the point (1, -2) to find the whole equation:
y−y0=m(x−x0)→y−(−2)=−1⋅(x−1)→y+2=−x+1→y=−x−1
In conclusion, the equations for the lines normal to the functions are:
1) y=−8x−851
2) y=−x−1
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