The line x = k intersects the graph of the parabola x = -y^2 - 3y - 5 at exactly one point. What is k?
What is k?
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\( x = -y^2 - 3y - 5\\ \frac{dx}{dy}=-2y-3=0\\ y_{max}=-\frac{3}{2}\\ x_{max}=k=-(-\frac{3}{2})^2-3\cdot(-\frac{3}{2})-5\\ \color{blue}k=-2.75\)
!
x = -y^2 - 3y -5
This is a parabola that opens to the left
The only way a line can intersect the parbola at one point is if it interesects it at the vertex
The y coodinate of the vertex is - (-3) / (2 * -1) = 3/2
So the x coordinate of the vertex = k is
-(3/2)^2 - 3 (3/2) - 5
-9/4 + 9/2 - 20/4 =
9/4 - 20/4 =
-11/4 = -2.75 = k