Leilani's Post

Bellringer: Find the axis of Symmetry

*Remember: Ax2+Bx+C

Formula to find the axis is:

x= -b/2(a)

a) y= x2+4x+1

If we apply the formula our equation would be:
x= -4/2(1) = -2

So.... x= -2

b) y= 2x2-6x+3

Equation: x= 6/2(2) =3/2

So... x= 3/2

c) y= -2x2+5x+1

Equation: x= -5/2(-2) =-5/-4

So...x= 5/4

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Practice Question 1:

y= 1x2+4x+1

1) Axis of symmetry

Equation: x= -4/2(1) =-4/2 =-2
So....x=-2

2) Y-intercept
The y-intercept is always C.

So our y-intercept is 1.

3)Vertex
Replace the x in the equation with your axis of symmetry, in our case it would be -2.

-2^2+4(-2)+1 =
4-8+1 = 4-8 = -4...........-4+1 = -3

So our vertex is (-2,-3)

4)Find the Zeroes of the equation
Solution of a quadtratic equation formula= -b+-{b2--4ac/2a

{ = square root

Type it into your calculator like so:
-4+({4^2-4*1*1))/2*1 2 = -4+({12))/2= -.27

Do it again for the negatively:
-4-({4^2-4*1*1))/2*1 2 = -4-({12))/2= -3.7

So our zeroes are: (-.27 , 0) and (-3.7 , 0)



Find the axis of:
a) 6x^2+2x+9

b) -5x^2-7x+3

Yay!