well hmmm does the graph open upwards or downwards? well
is a quadratic, if the leading term's coefficient is negative, Down, if positive, Up
a)
now, let's see the leading term, -x², what's its coefficient? is -1 * x² or -1x², is -1, so is negative, thus is opening downwards
c)
x-intercepts occur, when y = 0, namely y = -x²-4x-3, so setting y to 0
0 = -x² -4x -3 take a common factor of -1, thus
0 = -1 (x²+4x+3) <--- now, if we factor that out, notice, surely you've done many of these by now, so we end up with
[tex]\bf \begin{array}{lcclll}
0=x^2&+4x&+3\\
&\uparrow &\uparrow \\
&3+1&3\cdot 1
\end{array}\\\\
-----------------------------\\\\
0=(x+3)(x+1)\implies
\begin{cases}
0=x+3\implies &-3=x\\
0=x+1\implies &-1=x
\end{cases}[/tex]
so, the x-intercepts are at -3 and -1
d)
now, the y-intercepts, just set x = 0
y = -x²-4x-3, settting x to 0 y = -0²-4(0)-3, which is y = -3
so the sole y-intercept is at y = -3
now, let's get on to b)
b) [tex]\bf \textit{vertex of a parabola}\\\\
\begin{array}{lccclll}
f(x)=&-1x^2&-4x&-3\\
&\uparrow &\uparrow &\uparrow \\
&a&b&c
\end{array}\qquad
\left(-\cfrac{{{ b}}}{2{{ a}}}\quad ,\quad {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}\right)[/tex]
and those are the coordinates, notice a = -1, b = -4 and c = -3
now, for
e)
well, all you have to do is, once you have the vertex, pick an x-value on the left-hand-side of the vertex, get they value for "y", or OUTPUT,
then pick another x-value, on the right-hand-side of the vertex, get the "y" value again, and plot away
