For each quadratic relation, determine the y-intercept and the axis of symmetry
a) y= x^2 + 2
b) y= -x^2 -4

- shelby1290

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- shelby1290

For part A) I got y=2 because I set x to 0 to get my y-intercept
y=x^2+2
y=0^2+2
y=2 (0,2)

- shelby1290

I then set my y-int to 0 to get my x
y=x^2+2
0=x^2+2
0=1x+2
x=2 (2,0)

- shelby1290

Am I doing it right so far?
How do i find the axis of symmetry?

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## More answers

- anonymous

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- anonymous

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- shelby1290

Do I need to find all of the variables?

- anonymous

you can do it analytically, but with the vertex form it is not so easy to see it right away unless you are pretty well-versed with quadratic equations
with given vertex-form you can solve for the axis of symmetry by first converting the equation into the standard form \(f(x) = ax^2+bx+c\) and the formula for the axis of symmetry is \(\large x=\frac{-b}{2a} \) now you will notice that symmetry pertains to the relationship of one half to the other half of the graph and their similarity

- anonymous

if you looked at your other notes, you will probably also notice that the axis of symmetry is your x-value in your vertex

- anonymous

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- anonymous

the neat thing about it in vertex form is that it is already laid out to you where your axis of symmetry is.

- anonymous

but if you want to just punish yourself by doing more work, let us see if it is truly the formula I gave you for the axis of symmetry based on the standard form of quadratic equation
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- anonymous

let us provide a clear understanding of what x-intercept and y-intercepts are
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- shelby1290

Okay I'm following......

- anonymous

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- shelby1290

mhm

- anonymous

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- anonymous

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- anonymous

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- anonymous

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- anonymous

now that we have that clear
note that your axis of symmetry is not the the same as your intercepts
they may have the same value on some circumstances, but not always not the case

- anonymous

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- shelby1290

So, can the axis of symmetry basically be thought of as a midpoint?

- anonymous

somewhat but not quite because midpoint is a very precise definition

- anonymous

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- anonymous

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- anonymous

difficult and not possible, correct?

- shelby1290

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- anonymous

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- shelby1290

But it can kinda be anywhere around the perimeter of the circle...we'll just need coordinates

- shelby1290

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- shelby1290

Wait. I'm not sure if i made a mistake and labelled the vertex instead of the midpoint

- anonymous

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