Ans: (a) $b=2$, $3$ and $-5$ (b) $k>-16$ (c) $y=-16$
- Substitute $(4,9)$ into $f(x)$, we have
$\begin{array}{rcl}
(4)^2 +b(4) -15 & = & 9 \\
16 +4b – 15 & = & 9 \\
4b & = & 8 \\
b & = & 2
\end{array}$Substitute $f(x)=0$, we have
$\begin{array}{rcl}
x^2 +2x -15 & = & 0 \\
(x+5)(x-3) & = & 0 \\
\end{array}$Therefore, $x=-5$ or $x=3$. Hence, the two $x$-intercepts are $3$ and $-5$.
- Consider the equation,
$\begin{array}{rcl}
f(x) & = & k \\
x^2 +2x -15 & = & k \\
x^2 +2x -15 – k & = & 0
\end{array}$If the equation $f(x)=k$ has two distinct real roots, then we have
$\begin{array}{rcl}
\Delta & > & 0 \\
(2)^2 -4(1)(-15 – k) & > & 0 \\
4 +60 + 4k & > & 0 \\
4k & > & -64 \\
k & > & -16
\end{array}$ - Consider the equation $f(x)=k$. If $k=-16$, then $\Delta =0$. Then $f(x)=k$ has two equal real roots. i.e. the graph of $y=f(x)$ and the graph of $y=k$ intersect each other at one point. Therefore, $y=-16$ intersects the graph of $y=f(x)$ at only one point.
