Answer:
1) To find the $y-$intercept, we set $x=0$ in the equation:
\begin{align*}
y &= x^{2}-6x-16 \\
y &= 0^{2}-6(0)-16 \\
y &= -16
\end{align*}
Therefore, the $y$-intercept is $(0,-16)$.
2) To find the $x$-intercepts, we set $y=0$ in the equation and solve for $x$:
\begin{align*}
y &= (3x+2)(x-5) \\
0 &= (3x+2)(x-5) \\
\end{align*}
Using the zero product property, we have:
\begin{align*}
3x+2 &= 0 \quad \text{or} \quad x-5=0 \\
x &= -\frac{2}{3} \quad \text{or} \quad x=5\\
\end{align*}
Therefore, the $x$-intercepts are $(-\frac{2}{3},0)$ and $(5,0)$.
3) If a quadratic function written in standard form $y=a x^{2}+bx+c$ has a negative $a$ parameter, then the parabola opens downwards.
4) To find the $x$-intercepts, we set $y=0$ in the equation and solve for $x$:
\begin{align*}
y &= x^{2}+4x-21 \\
0 &= x^{2}+4x-21 \\
\end{align*}
Using factoring or the quadratic formula, we get:
\begin{align*}
(x+7)(x-3) &= 0 \\
x &= -7 \quad \text{or} \quad x=3 \\
\end{align*}
Therefore, the $x$-intercepts are $(-7,0)$ and $(3,0)$.
To find the $y$-intercept, we set $x=0$ in the equation:
\begin{align*}
y &= 0^{2}+4(0)-21 \\
y &= -21
\end{align*}
Therefore, the $y$-intercept is $(0,-21)$.
Explanation:
The answer is in the picture.