7)
f(x) = 2x² - 12x + 22
now, if you recall, in a perfect square trinomial, the middle term is a combination of the two other terms times 2, namely
[tex]\bf \begin{array}{cccccllllll}
{{ a}}^2& + &2{{ a}}{{ b}}&+&{{ b}}^2\\
\downarrow && &&\downarrow \\
{{ a}}&& &&{{ b}}\\
&\to &({{ a}} + {{ b}})^2&\leftarrow
\end{array}\qquad
% perfect square trinomial, negative middle term
\begin{array}{cccccllllll}
{{ a}}^2& - &2{{ a}}{{ b}}&+&{{ b}}^2\\
\downarrow && &&\downarrow \\
{{ a}}&& &&{{ b}}\\
&\to &({{ a}} - {{ b}})^2&\leftarrow
\end{array}[/tex]
so let's us first do some grouping,
f(x) = (2x² - 12x) +22
f(x) = 2(x² - 6x) + 22
f(x) = 2(x² - 6x + [?]²) + 22
notice, we're missing a 3rd element in the group, to get a perfect square trinomial, hmmm what would that be?
well, let's see the middle guy, 6x, recall the middle guy is a combination of 2*a * b, therefore the middle term is 2 * x * b, and we know is 6x, thus
[tex]\bf2x\boxed{?}=6x\implies \boxed{?}=\cfrac{6x}{2x}\implies \boxed{?}=3[/tex]
low and behold, our missing fellow is 3 then.
now, keep in mind that, if we add 3², we also have to subtract 3², since all we're doing is borrowing from our very good friend Mr Zero, 0, thus
f(x) = 2(x² - 6x +3² - 3²) + 22
f(x) = 2(x² - 6x +3²) + 22 -2*3² <---- notice the common factor 2 there.
f(x) = 2(x - 3)² + 22 - 18
f(x) = 2(x - 3)² +4
9)
[tex]\bf ~~~~~~\textit{parabola vertex form}
\\\\
\begin{array}{llll}
y=a(x-{{ h}})^2+{{ k}}\\\\
x=a(y-{{ k}})^2+{{ h}}
\end{array}
\qquad\qquad
vertex~~(\stackrel{}{{{ h}}},\stackrel{}{{{ k}}})\\\\
-------------------------------\\\\
h(x)=5(x+1)^2-45\implies h(x)=5[x-\stackrel{h}{(-1)}]^2\stackrel{k}{-45}[/tex]
11)
we know the roots are -2 and 3, therefore then, recall that a "solution" or a zero is really just referring to the x-intercepts, so -2 and 3 are really where the graph touches the x-axis, or the x-intercepts.
now, if x = -2, then x + 2 = 0 <--- this means (x+2) is a factor
if x = 3, then x - 3 = 0 <--- this means (x - 3) is a factor as well.
now if could have some other factors, like say 5, and that's fine, but we know that, since -2 and 3 are zeros, (x+2)(x-3) are part of it, so 5(x+2)(x-3) is very feasible.
