Find the equation of the tangent to the curve `y- x^2 + 4x - 5 = 0`   at the point (5,10) y-X2+4x-5=o

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 Write the equation in standard form (y=)

`y= x^2 -4x + 5` To find the tangent-find the gradient at the point (5;10)   

`y'=2x - 4`  (Remember, the derivative gives you the gradient. Substitute for x as we want the gradient at that point.)

`y' = 2(5) - 4`

`y'...

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 Write the equation in standard form (y=)

`y= x^2 -4x + 5` To find the tangent-find the gradient at the point (5;10)   

`y'=2x - 4`  (Remember, the derivative gives you the gradient. Substitute for x as we want the gradient at that point.)

`y' = 2(5) - 4`

`y' = 6` = m (the gradient)

A tangent has an equation : y=mx+ c

We have a point (5;10) and we have m and can therefore find c:

y= mx+c becomes

`10 = (6)(5) +c`

`therefore c=-20`

`therefore` y=6x - 20

 

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