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# Find the point on the curve y is equal to x raised to 3 minus 11 x plus 5 at which the tangent is y is equal to x minus 11.

9) Find the point on the curve $y = x^3 - 11x + 5$ at which the tangent is $y = x -11$

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We know that the  equation of a line is y = mx + c
Know the given equation of tangent is
y = x - 11
So, by comparing with the standard equation we can say that the slope of the tangent (m) = 1 and value of c if -11
As we know that  slope of the tangent at a point on the given curve is given by  $\frac{dy}{dx}$
Given the equation of curve is
$y = x^3 - 11x + 5$
$\frac{dy}{dx} = 3x^2 -11$
$3x^2 -11 = 1\\ 3x^2 = 12 \\ x^2 = 4 \\ x = \pm2$
When x = 2 , $y = 2^3 - 11(2) +5 = 8 - 22+5=-9$
and
When x = -2 , $y = (-2)^3 - 11(22) +5 = -8 + 22+5=19$
Hence, the coordinates are (2,-9) and (-2,19), here (-2,19) does not satisfy the equation y=x-11

Hence, the coordinate is (2,-9) at which the tangent is $y = x -11$

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