Persamaan garis singgung kurva y=x^2+4x-3 dititik (1,2)

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Persamaan garis singgung kurva y=x^2+4x-3 dititik (1,2)

PGS di titik (1,2)
y= m( x – 1) + 2 = mx + 2 – m
Subt pers garis ke pers kurva

y=x²+4x-3
y= 
(mx + 2 – m)² + 4(mx + 2 – m) -3
  = m²x² + 2 mx(2-m) + 4 – 4m + m² + 4mx + 8 – 4m -3
  = m²x² + 4 mx – 2m²x + 4 mx + m² -4m -4m + 4 + 8 – 3
  = m²x² + (-2m² + 8m) x + (m² – 8m + 9 ) = 0
Menyinggung Sy. D = 0
⇒ (
-2m² + 8m)² – 4 (m²)(m² -8m +9)                     = 0
=   4(m²)² – 32m³ + 64m² – 4(m²)² + 32m³ – 36m²  = 0
=                                                  64 m² – 36 m²   = 0
                                                                         m = 0
PGS =

y = (0)( x – 1) + 2
y = 2