Question

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?

• A. $-4\sqrt{2}$
• B. $-2-2\sqrt{2}$
• C. $-2$
• D. $-\sqrt{2}$
• E. $-2+\sqrt{2}$

Answer 1

Answer: (B)

$-2-2\sqrt{2}$

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

If $y=x^2-2x+7$ is tangent to $y=kx+5$

then they intersect $kx+5=x^2-2x+7$

If tangent then diterminant is zero

$(2+k{)}^2-8=0k=\pm 2\sqrt{2}-2$

Negative value of $k$ is $-2\sqrt{2}-2$