## Precalculus (10th Edition)

$a=\pm3$
If $f\circ g$ crosses the y-axis at $23$ that means that $(f\circ g)(0)=23.$ $g(0)=3(0)+a=0$. Thus $(f\circ g)(0)=f(g(0))=f(a)=2a^2+5=23.$ Solve the above equation to obtain: $2a^2+5=23\\2a^2=18\\a^2=9\\a=\pm3$