Question

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

Answer 1

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`

Answer 2

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).