your text should have the following:
for y = a(x-p)^2 + q
vertex is (p,q)
axis of symmetry: x=p
opening: up if a>0, down if a<0
please redo your answers and let me know what you got.
Identify the vertex, axis of symmetry, and direction of opening for y=2(x+3)^2-5.
vertex = 1/2
axis = 1/2
direction = up
3 answers
y=2x+3^2-5
y=2x+9-5
x=-b/2a
x=-9/2(-9)
x=1/2
I don't know if I have the value for a right
y=2x+9-5
x=-b/2a
x=-9/2(-9)
x=1/2
I don't know if I have the value for a right
vertex= (3,-5)
axis of symmetry= x=2
direction= up b/c 2>0
axis of symmetry= x=2
direction= up b/c 2>0