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Find the equation of the tangent to: y = X^2 + 3x + 2 at the point (2,12)

(1) Find the gradient using differentiation (2) If the gradient at (x₁,y₁) is m,y - y₁ = m(x - x₁)
(1) We differentiate the given equation:dy/dx = 2x + 3
Then, find the gradient at (2,12). Sub x= 2 into dy/dx = 2x + 3 dy/dx = 2(2) + 3dy/dx = 7
(2) y-12 = 7(x-2) y-12=7x-14 y=7x-2

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Find the equation of the tangent to: y = X^2 + 3x + 2 at the point (2,12)

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