quadratic graphs 2

$y=-\left(x-3\right)^2-1$

$y=\left(x-3\right)^2-1$

$y=-\left(x+3\right)^2-1$

$y=-\left(x-1\right)^2-3$

$y=-\left(x-3\right)^2+1$

 $y=-6x^2$  

 $y=-x^2+6$  

 $y=-\left(x+6\right)^2$  

 $y=-\left(x-6\right)^2$  

 $y=-x^2-6$  

A parabola must have two x-intercepts.

A parabola can have two, one or no x-intercepts.

Some parabolas do not have a y-intercept.

Some parabolas do not have a turning point.

The axis of symmetry is half way between the x-intercept and the y-intercept.

 $\left(0,-4\right)$  

 $\left(0,-3\right)$  

 $\left(0,\ 1\right)$  

 $\left(3,0\right)$  

 $\left(-1,0\right)$  

 $y=x^2-4$  

 $y=-x^2+4$  

 $y=-\left(x-4\right)^2$  

 $y=-\left(x+4\right)^2$  

 $y=-x^2-4$  

Vertex (3,8) Axis x = 3

Vertex (3,8) Axis y = 3

Vertex (8,3) Axis x = 3

Vertex (8,3) Axis y = 3

roots or x-intercepts

parabola

axis of symmetry

y intercept