Question

What are the coordinates of the vertex for f(x) = x2 + 6x + 13? Walk me through the steps please?

Answer 1

I can rewrite your function as below:
\[\begin{gathered}
y = {x^2} + 6x + 9 - 9 + 13 \hfill \\
y = {\left( {x + 3} \right)^2} + 4 \hfill \\
y - 4 = {\left( {x + 3} \right)^2} \hfill \\
\end{gathered} \]

Answer 2

where did the 9's come from?

Answer 3

the 4 too

Answer 4

It is a my choice in order to get a perfect square: \((x+3)^2\)

Answer 5

oh i see!

Answer 6

\(4\) comes from \(-9+13=4\)
In other words I added and subtracted \(9\), and such operation doesn't change the right side of your equation

Answer 7

Oh okay. Thank you very much!

Answer 8

next step?

Answer 9

now I maje this traslation:
\[\left\{ \begin{gathered}
Y = y - 4 \hfill \\
X = x + 3 \hfill \\
\end{gathered} \right.\]

Answer 10

oops.. I made*...

Answer 11

gotcha

Answer 12

sorry.. I make*

Answer 13

where \(Y,X\) are the new coordinates
Using those new coordinates our equation can be rewritten as below:
\[Y = {X^2}\]

Answer 14

These are my answer choices:

A. 2(x + 5)(x − 4)

B. 2(x − 5)(x + 4)

C. 2(x + 5)^2

D. 2(x + 5)(x − 5)

Answer 15

I think that those options are related to another exercise

Answer 16

oh I'm sorry! you're right. gimme a sec.

Answer 17

no problem :)

Answer 18

the coordinates of the vertex of this parabola \(Y=X^2\) are:
\(Y=0,X=0\)

Answer 19

substituting \(X=0,Y=0\) into the equation of my trasformation, I get:
\[\left\{ \begin{gathered}
y - 4 = 0 \hfill \\
x + 3 = 0 \hfill \\
\end{gathered} \right.\]
please find \(x,y\)

Answer 20

Here are my answer choices for real:

A. (4, 4)

B. (−4, 4)

C. (3, 4)

D. (−3, 4)

and okay will do

Answer 21

for example, from the first equation, I get \(y=4\), and from the second equation, we get: \(x=...?\)

Answer 22

x = -3

Answer 23

correct! so what is the right option?

Answer 24

D. (-3, 4)!

Answer 25

that's right!

Answer 26

Yes! Thank you for your help!