A straight line passes through the point (2, 0) and has gradient m. Find the two values of m for which the line is a tangent to the curve y = x^2 − 4x + 5. For each value of m, find the coordinates of the point where the line touches the curve.

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`y= x^2-4x+5`

The gradient(n) of the tangent for the curve is given by;

`dy/dx = 2x-4`

`n=2x-4`

The equation of line passing (2,0) is given by;

`y = mx+c`

Since the line passes through (2,0)

`0 = 2xm+c`

`c = -2m`

The equation of the line is;

`y = mx-2m`

If the line is tangent to the curve it satisfies the gradient of tangent.

`y=(2x-4)x-2(2x-4)`

`y=2x^2-4x-4x+8=2x^2-8x+2`

At the tangent point

`2x^2-8x+8 = x^2-4x+5`

`x^2-4x+3 = 0`

`x^2-3x-x+3=0`

`x(x-3)-1(x-3)=0`

`(x-3)(x-1)=0`

`x=3` and `x=1`

When `x=3` then `m=2xx3-4=2`

When `x=1` then `m=2xx1-4=-2`

When x = 3 then `y = 2^2 − 4xx2 + 5 = 1`

When `x = 1` then `y = 1^2 − 4xx1 + 5 = 2`

*So the two values of in which the straight line is a tangent to curve is m=2 and m=-2.*

*The curve touches the line at points (3,1) and (1,2).*

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