Question

The maximum vertical distance $d$ between the parabola $y=-2x^2+4x+3$ and the line $y=x-2$ throughout the bounded region in the figure, is

• A. $\frac{47}{8}$
• B. $\frac{49}{8}$
• C. $\frac{50}{8}$
• D. $\frac{48}{8}$

Answer 1

Answer: (B)

The vertical distance is given by
$d=-2x^2+4x+3-(x-2)=-2x^2+3x+5$, which is a parabola opening down.
Its maximum value is the $y$-coordinate of the vertex which has $x$-coordinate equal to
$\frac{-b}{2a}=\frac{-3}{2(-2)}=\frac{3}{4}$. The y-coordinate, then, is $-2{\left(\frac{3}{4}\right)}^2+3\left(\frac{3}{4}\right)+5=\frac{-9}{8}+\frac{18}{8}+\frac{40}{8}=\frac{49}{8}$