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= 2x²-4x-x+2
= 2x²-5x+2
Vemos que el punto p es (-2,f(-2))
Reemplazamos -2
f(-2)= 2×(-2)²-5×(-2)+2
= 2×4+10+2=
=20
p= (-2,20)
Derivamos f(x)=2x²-5x+2
quedando:
f'(x)= 2×2x -5 = 4x -5
Reemplazamos en x=-2
f'(-2)= 4×(-2)-5 = -8-5
= -13
Luego la pendiente de la recta tangente es -13
Ocupamos la formula punto-pendiente:
y-20=-13(x-(-2))
y-20=-13(x+2)
y= -13x -26+20
y=-13x -6