Question

Find an equation for the tangent to the curve at the given point. y=x^2+2 ; (2, 6)

Answer 1

1) Find the derivative $y'=2x$ and $y'(2)=2\cdot2=4$.
2) The equation of the tangent line is $y=y'(x_0)(x-x_0)+y_0$, so at point (2,6): $x_0=2 \\ y_0=6$ and
$y=4(x-2)+6$
$y=4x-2$ -the equation of the tangent line at point (2,6).

Answer 2

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

$\bf y=x^2+2\implies \left. \cfrac{dy}{dx}=2x \right|_{x=2}\implies 2(2)\implies \stackrel{m}{\underline{4}} \\\\\\ \stackrel{\textit{point-slope form}}{y- y_1= m(x- x_1)}\implies y-6=\underline{4}(x-2) \\\\\\ y-6=4x-8\implies y=4x-2$