<p>Find a negative value of k so that the graph of y = x^2 - 2x + 7 and the graph of y = kx + 5 are tangent?</p>
<p>A) - 4√2
B) - 2 - 2√2
C) - 2
D) - √2
E) - 2 + √2</p>
<p>choose B…</p>
<p>Take the first derivative of the 1st equation and equate to the 2nd equation to get the tangent of the parabola. Sounds like (c).</p>
<p>For the point of tangency,
y=x^2-2x+7
dy/dx=2x-2=k (Slope of y=kx+5)
And kx+5=x^2-2x+7
(2x-2)x+5=x^2-2x+7
2x^2-2x+5=x^2-2x+7
x^2-2=0
x=sqrt 2 or -sqrt 2
k=2 sqrt 2 -2>0 (rejected) or k=-2 sqrt 2 - 2</p>
<p>B</p>
<p>I agree with the answer given be warsovreign, B, but I’m surprised to see a problem requiring calculus knowledge, though basic, on the Math II subject test. Is there another, non-calculus way to do this problem apart from plugging in the answers and graphing?</p>
<p>Well, there can be a non-calculus solution.
At the point of tangency
x^2-2x+7=kx+5
x^2-(k+2)x+2=0
As y=x^2-2x+7 and y=kx+5 touches at one point,
delta=0
(k+2)^2-4(1)(2)=0
k+2= 士 2 sqrt 2
k=-2 sqrt 2 - 2<0</p>