Find the equation of the tangent to the graph of f(x)=x^2+1 at the point (2, 5) using limits.

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justaguide | College Teacher | (Level 2) Distinguished Educator

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The equation of the tangent to the the curve given by the function y = f(x) at any point `(x', f(x'))` is `(y - f(x'))/(x - x') = lim_(h->0)(f(x' + h) - f(x'))/h`

For f(x) = 1 + x^2 the equation of the tangent at the point (2, 5) is

`(y - 5)/(x - 2) = lim_(h->0)(f(2 + h) - f(2))/h`

=> `(y - 5)/(x - 2) = lim_(h->0)((2 + h)^2 + 1 - 2^2 - 1)/h`

=> `(y - 5)/(x - 2) = lim_(h->0)(4 + h^2 + 4h + 1 - 4 - 1)/h`

=> `(y - 5)/(x - 2) = lim_(h->0)(h + 4)`

Substitute h = 0

=> `(y - 5)/(x - 2) = 4`

=> `y - 5 = 4x - 8`

=> 4x - y - 3 = 0

The equation of the tangent to y = x^2 + 1 at the point (2, 5) is 4x - y - 3 = 0

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