Question

Write the quadratic function in the form g(x)= a (x-h)^2 +k . Then, give the vertex of its graph. g(x)=3x^2+30x+72
Writing in the form specified: g(x)= ___
Vertex:___

Answer 1

so hmmm as you may know by now...   $\bf \begin{array}{cccccllllll} {{ a}}^2& + &2{{ a}}{{ b}}&+&{{ b}}^2\\ \downarrow && &&\downarrow \\ {{ a}}&& &&{{ b}}\\ &\to &({{ a}} + {{ b}})^2&\leftarrow \end{array}\qquad % perfect square trinomial, negative middle term \begin{array}{cccccllllll} {{ a}}^2& - &2{{ a}}{{ b}}&+&{{ b}}^2\\ \downarrow && &&\downarrow \\ {{ a}}&& &&{{ b}}\\ &\to &({{ a}} - {{ b}})^2&\leftarrow \end{array}$

is a perfect square trinomial

now, let's take a closer look at the middle guy
is just 2 * guy on the left * guy on the right
so the middle term is just the product of the  two terms of the binomial, without the exponent, or square rooted, and 2

alrite... that said.... now let's do some grouping

$\bf g(x)=3x^2+30x+72\implies g(x)=(3x^2+30x)+72 \\\\\\ g(x)=3(x^2+10x)+72\implies g(x)=3(x^2+10x+\boxed{?}^2)+72$

so hmm we're missing the guy on the right there... who may that be?

well, we know the middle term is 10x, wait a minute, 10x is just 2 * 5 * x

so, if the guy on the left is "x", the other guy must be "5", 2*5*x = 10x

now, what we do is, we add 5²

however, bear in mind that, all we're doing is, borrowing from our very good friend Mr Zero, 0

so, if we add 5², we also have to subtract 5², so +5² - 5²

let's do so

$\bf g(x)=3(x^2+10x)+72\implies g(x)=3(x^2+10x\boxed{+5^2-5^2})+72 \\\\\\ \textit{let's take out the negative from the group} \\\\\\ g(x)=3(x^2+10x+5^2)+\boxed{3(-5^2)}+72 \\\\\\ g(x)=3(x^2+10x+5^2)-75+72\implies g(x)=3(x+5)^2-3 \\\\\\ g(x)=3[x-(-5)]^2-3$

$\bf \qquad \textit{parabola vertex form}\\\\ \begin{array}{llll} y=a(x-{{ h}})^2+{{ k}} \end{array} \qquad\qquad vertex\ ({{ h}},{{ k}})\\\\ -------------------------------\\\\ \begin{array}{lcclll} g(x)=3[x-&(-5)]^2&-3\\ &\uparrow &\uparrow \\ &h&k \end{array}$