Deriving Quadratic Functions by Virounika Mina https://padlet.com/1026739/c7xosngsm854 Made with a taste for adventure en-us 2019-02-15 17:20:25 UTC 2019-02-16 01:40:27 UTC hello@padlet.com Deriving Quadratic Functions 1026739 https://padlet.com/1026739/c7xosngsm854/wish/331829542 Given: Vertex (5,9); point (3. -3)]]> 2019-02-15 17:26:11 UTC https://padlet.com/1026739/c7xosngsm854/wish/331829542 Putting into Vertex Form 1026739 https://padlet.com/1026739/c7xosngsm854/wish/331832802 f(x)= a(x-h)2-k
f(x)= a(x-5)2+9
You begin by plugging in your vertex point into vertex form. It is important to change your x-coordinate's sign, which is why you go from 5 to -5 in this case. ]]> 2019-02-15 17:33:43 UTC https://padlet.com/1026739/c7xosngsm854/wish/331832802 Solving for a 1026739 https://padlet.com/1026739/c7xosngsm854/wish/331835463 additional point: (3, -3)
f(x)= a(x-5)2+9
y= a(x-h)2+9
-3= a(3-5)2+9
-3= 4a+9
-9      -9
a= -3
1. Look at the additional point provided and plug it in
2. Change f(x)= to y=
3. you now have the variables x and y in your equation, which is where you plug in the 3 of the other point and the -2 (3 for x and -2 for y)
4. Using the order of operations subtract in the parenthesis and square it, then multiply by a and add 9 to get 4a.
5. Divide to isolate a and a equals -3
6. After solving for what a equals, put -3 in for the a in the vertex form ]]> 2019-02-15 17:39:33 UTC https://padlet.com/1026739/c7xosngsm854/wish/331835463 Answer 1026739 https://padlet.com/1026739/c7xosngsm854/wish/331838925 f(x)= -3(x-5)2+9]]> 2019-02-15 17:47:05 UTC https://padlet.com/1026739/c7xosngsm854/wish/331838925