Find the equation of the tangent to the curve y=2x^2+3 that pass through point (2,-7)

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You need to use the formula that gives the equation of the tangent line to a curve, at a point,`(x_0,y_0)` , such that:

`f(x) - f(x_0) = f'(x_0)(x - x_0)`

The problem provides the coordinates `(x_0,f(x_0)) = (2,-7)` and you need to evaluate `f'(x)` at `x_0 = 2` , such that:

`f'(x) = 4x => f'(2) = 4*2 => f'(2) = 8`

`f(x) - (-7) = 8(x - 2) => f(x) + 7 = 8x - 16`

Substituting y for f(x) yields:

`8x - y - 16 - 7 = 0 => 8x - y - 23 = 0`

**Hence, evaluating the equation of the tangent line to the curve, at the point `(2,-7)` yields **`8x - y - 23 = 0.`

y=`2x^2 + 3`

y=`x^(n)` `dy/dx=nx^(n-1)`

`dy/dx` =`4x`

(2,-7)

gradient = 4(2)

=8

`y=mx+c`

`m=8`

`y-y_(o)=m(x-x_(o))`

`Let x_(o) = 2 and y_(o) =-7`

`y-(-7)=8(x-2)`

`y+7=8x-16`

`y=8x-23`

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