shelby1290
  • shelby1290
For each quadratic relation, determine the y-intercept and the axis of symmetry a) y= x^2 + 2 b) y= -x^2 -4
Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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shelby1290
  • 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
  • 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
  • shelby1290
Am I doing it right so far? How do i find the axis of symmetry?

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anonymous
  • anonymous
|dw:1441945461825:dw|
anonymous
  • anonymous
|dw:1441945660612:dw|
shelby1290
  • shelby1290
Do I need to find all of the variables?
anonymous
  • 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
  • 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
  • anonymous
|dw:1441946631528:dw|
anonymous
  • 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
  • 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 |dw:1441946825142:dw|
anonymous
  • anonymous
let us provide a clear understanding of what x-intercept and y-intercepts are |dw:1441947158858:dw|
shelby1290
  • shelby1290
Okay I'm following......
anonymous
  • anonymous
|dw:1441947258243:dw|
shelby1290
  • shelby1290
mhm
anonymous
  • anonymous
|dw:1441947409829:dw|
anonymous
  • anonymous
|dw:1441947459955:dw|
anonymous
  • anonymous
|dw:1441947486217:dw|
anonymous
  • anonymous
|dw:1441947550094:dw|
anonymous
  • 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
  • anonymous
|dw:1441947737731:dw|
shelby1290
  • shelby1290
So, can the axis of symmetry basically be thought of as a midpoint?
anonymous
  • anonymous
somewhat but not quite because midpoint is a very precise definition
anonymous
  • anonymous
|dw:1441947959255:dw|
anonymous
  • anonymous
|dw:1441947998746:dw|
anonymous
  • anonymous
difficult and not possible, correct?
shelby1290
  • shelby1290
|dw:1441948105567:dw|
anonymous
  • anonymous
|dw:1441948201968:dw|
shelby1290
  • shelby1290
But it can kinda be anywhere around the perimeter of the circle...we'll just need coordinates
shelby1290
  • shelby1290
|dw:1441948232996:dw|
shelby1290
  • shelby1290
Wait. I'm not sure if i made a mistake and labelled the vertex instead of the midpoint
anonymous
  • anonymous
|dw:1441948267352:dw|
shelby1290
  • shelby1290
|dw:1441948324139:dw|