Find a negative value of k so that the graph of y = x2 - 2x + 7 and the graph of y = kx + 5 are tangent

Show work please, I am studying for the Sat II Math II
I am not sure how I can solve this.

y = x² - 2x + 7
y = kx + 5

If the curves are tangent, their slopes are equal at the point of tangency:
y' = 2x-2 = k

If the curves are tangent, the point of tangency is common to both curves:
x² - 2x + 7 = y = kx + 5
x² - 2x + 7 = kx + 5 = (2x-2)x + 5
x² - 2x + 7 = 2x² - 2x + 5
x² - 2 = 0
x = ±√2
k = ±2√2 - 2 = 2(±√2 - 1)

If you need k < 0, k = -2(√2+1)

If you need k > 0, k = 2(√2-1)

Here'a a solution to your problem...

https://docs.google.com/viewer?a=v&pid=e…

Hope this helps

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