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shelby1290
 one year ago
For each quadratic relation, determine the yintercept and the axis of symmetry
a) y= x^2 + 2
b) y= x^2 4
shelby1290
 one year ago
For each quadratic relation, determine the yintercept and the axis of symmetry a) y= x^2 + 2 b) y= x^2 4

This Question is Closed

shelby1290
 one year ago
Best ResponseYou've already chosen the best response.0For part A) I got y=2 because I set x to 0 to get my yintercept y=x^2+2 y=0^2+2 y=2 (0,2)

shelby1290
 one year ago
Best ResponseYou've already chosen the best response.0I then set my yint to 0 to get my x y=x^2+2 0=x^2+2 0=1x+2 x=2 (2,0)

shelby1290
 one year ago
Best ResponseYou've already chosen the best response.0Am I doing it right so far? How do i find the axis of symmetry?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1441945461825:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1441945660612:dw

shelby1290
 one year ago
Best ResponseYou've already chosen the best response.0Do I need to find all of the variables?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you can do it analytically, but with the vertex form it is not so easy to see it right away unless you are pretty wellversed with quadratic equations with given vertexform 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
 one year ago
Best ResponseYou've already chosen the best response.0if you looked at your other notes, you will probably also notice that the axis of symmetry is your xvalue in your vertex

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1441946631528:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the neat thing about it in vertex form is that it is already laid out to you where your axis of symmetry is.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but 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
 one year ago
Best ResponseYou've already chosen the best response.0let us provide a clear understanding of what xintercept and yintercepts are dw:1441947158858:dw

shelby1290
 one year ago
Best ResponseYou've already chosen the best response.0Okay I'm following......

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1441947258243:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1441947409829:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1441947459955:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1441947486217:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1441947550094:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0now 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
 one year ago
Best ResponseYou've already chosen the best response.0dw:1441947737731:dw

shelby1290
 one year ago
Best ResponseYou've already chosen the best response.0So, can the axis of symmetry basically be thought of as a midpoint?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0somewhat but not quite because midpoint is a very precise definition

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1441947959255:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1441947998746:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0difficult and not possible, correct?

shelby1290
 one year ago
Best ResponseYou've already chosen the best response.0dw:1441948105567:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1441948201968:dw

shelby1290
 one year ago
Best ResponseYou've already chosen the best response.0But it can kinda be anywhere around the perimeter of the circle...we'll just need coordinates

shelby1290
 one year ago
Best ResponseYou've already chosen the best response.0dw:1441948232996:dw