# Given `f(x)=5x^2-9x+11 ` find the equation of the tangent line at x=2 use `lim_(h->0)(f(a+h)-f(a))/h` with a=2

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The function `f(x)=5x^2-9x+11` . The slope of the tangent at a = 2 is equal to `f'(x)` at x = 2.

To determine `f'(2)` , the value of `lim_(h->0)(f(2 +h) - f(2))/h` has to be estimated.

`lim_(h->0)(f(2 +h) - f(2))/h`

=> `lim_(h->0)(5*(2 +h)^2-9*(2+h)+11 - 5*2^2+9*2-11)/h`

=> `lim_(h->0)(5*(4 + 4h + h^2)- 18 - 9h+11 - 20+18-11)/h`

=> `lim_(h->0)(20 + 20h + 5h^2- 18 - 9h+11 - 20+18-11)/h`

=> `lim_(h->0)(20h + 5h^2 - 9h)/h`

=> `lim_(h->0) 20 + 5h - 9`

substituting h = 0

=> 11

The slope of the tangent is 11

`f(2) = 20 - 18 + 11 = 13`

The equation of the tangent is `(y - 13)/(x - 2) = 11`

=> `y - 13 = 11x - 22`

=> `11x - y - 9 = 0`

**The required equation of the tangent is 11x - y - 9 = 0**