Respuesta :

bearing in mind that in (2, 6),  x₁ = 2,    y₁ = 6.

[tex]\bf y=x^2+2\implies \left. \cfrac{dy}{dx}=2x \right|_{x=2}\implies 2(2)\implies \stackrel{m}{\underline{4}} \\\\\\ \stackrel{\textit{point-slope form}}{y- y_1= m(x- x_1)}\implies y-6=\underline{4}(x-2) \\\\\\ y-6=4x-8\implies y=4x-2[/tex]
frika
1) Find the derivative [tex]y'=2x[/tex] and [tex]y'(2)=2\cdot2=4[/tex].
2) The equation of the tangent line is [tex]y=y'(x_0)(x-x_0)+y_0[/tex], so at point (2,6): [tex]x_0=2 \\ y_0=6[/tex] and
[tex]y=4(x-2)+6[/tex]
[tex]y=4x-2[/tex] -the equation of the tangent line at point (2,6).

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