Find the equation of the tangent to the curve `y- x^2 + 4x - 5 = 0` at the point (5,10) y-x^2+4x-5=0
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Write the equation in standard form:
`y= x^2 - 4x+ 5 ` Use the derivative to find the gradient (multiply the power of x by the coefficient so 2 x 1 = 2 and then minus 1 from the power so `x^2` becomes x and the same for -4x so 1x -4=-4 and the x power reduces to 0 `x^(1-1)` :
`y' = 2x - 4` now substitute the x value in terms of the question(5;10)
= 2(5) - 4
= 6
As a tangent has the equation y=mx + c now substitute the x value, y value and value of m into the new equation to solve for c:
y= mx+c
10= (6)(5) +c
10= 30 + c
-20 = c
`therefore` y=6x - 20
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