Question

Graph the equation and label the vertex and the axis of symmetry

Answer 1

\[y=x^2-1\]

Answer 2

Make a table of x values. Start graphing from -1.

Answer 3

does x=1?

Answer 4

and y=-1?

Answer 5

I don't know about that, sorry.

Answer 6

@terenzreignz

Answer 7

Oh... what do you know, I'm here already XD

Answer 8

haha that's a first! lol!

Answer 9

No it isn't, it isn't my fault, anyway...
So, you already know what the vertex is?

Answer 10

hmm, is it (1,-1)?

Answer 11

how do I find the vertex?

Answer 12

first things first.... When you have a quadratic function of the form
\[\Large f(x) = \color{blue}ax^2 + \color{red}bx + \color{green}c\]
Then the x-value of the vertex is given by
\[\Large x = -\frac{\color{red}b}{2\color{blue}a}\]

Answer 13

so is that 0/2(1)=0.... so x=0?

Answer 14

Yup.
and to get the y-value, simply get f(0)

Answer 15

y=0-1=-1

Answer 16

Actually \(\Large f\left(-\frac{\color{red}b}{2\color{blue}a}\right)\)

And it results (in general) in
\[\Large y = \frac{4\color{blue}a\color{green}c-\color{red}b^2}{4\color{blue}a}\]

So yeah, -1 is right.

And that's your vertex, (0,-1)

Answer 17

so the vertex is (0,-1) then.

Answer 18

That's right.
And the axis of symmetry is the line
x = <whatever is the x-value of your vertex>

So the axis of symmetry is...?

Answer 19

0

Answer 20

That's not an equation :P
Let's be formal here XD

Answer 21

x=0 lol

Answer 22

And there you have it, your vertex and axis of symmetry.
As for graphing...

Answer 23

Have you done that already?

Answer 24

from the vertx I go one up and to the left and right and then 3 up and to the left and right?

Answer 25

Doesn't matter. Just plot a few points and trace out the trend.

Answer 26

It should ultimately resemble the graph @ravikr.iit posted.

Answer 27

okay thanks again!

Answer 28

No problem ^_^

Answer 29

and happy pi day!

Answer 30

3/15 here already :P

So sleepy, I'm signing off now
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Terence out
^_^