Question

Please help me write the function in vertex form and identify its vertex. g(x) = x2 - 10x + 11

Answer 1

−b2a=102=5
and
g(5)=52−10×5+11=25−50+11=−25+11=−14
so vertex is
(5,−14)
and vertex form will be
g(x)=(x−5)2−14

Answer 2

is this correct?

Answer 3

http://www.purplemath.com/modules/sqrvertx.htm this is a good tutorial on vertex form. Sorry, I can't help much more tonight, but I have a calculus test tomorrow that I have to study for. I am sure somebody will check your work. Just use Khan academy or purplemath for tutorials. Khan academy has detailed videos and is free.

Answer 4

hmmmm

Answer 5

Okay, thank you sm for your hep today! Good luck on your test tomorrow

Answer 6

ahemm @SahiraAlvarado do you know how to "complete the square"?

Answer 7

No problem. I'm sure I will see you around open study. Nice meeting you.

Answer 8

You certainly will! I'll be starting my Algebra 2B online class shortly... And it was a pressure meeting you and thanks again for your assistance! It was extremely beneficial.

Answer 9

\(\bf x^2 - 10x + 11\implies (x^2 - 10x) + 11\implies (x^2 - 10x+{\color{red}{ \square }}^2) + 11\)

what do you think we need there, in the group, to get a "perfect square trinomial" ?

Answer 10

25? @jdoe0001

Answer 11

well, yes, \(\bf 5^2 \implies 25\) , keep in mind that all we're doing is borrowing from Zero 0, so if we ADD 25, we have to also SUBTRACT 25, as well

so \(\bf x^2 - 10x + 11\implies (x^2 - 10x) + 11\implies (x^2 - 10x+{\color{red}{ 5 }}^2) + 11-{\color{red}{ 5 }}^2
\\ \quad \\
(x-5)^2+11-25\implies g(x)=(x-5)^2-14\)

Answer 12

\(\large \begin{array}{llll}
g(x)=(x-&5)^2&-14\\
&\uparrow &\uparrow \\
\end{array}\bf \qquad vertex\implies (5,-14)\)

Answer 13

I got it!!!
Thanks