The vertex is (0,-8)
im not sure about axis of symmetry but i don’t think it has one
Answer:
B'(16,14)
Step-by-step explanation:
First find the coordinates of the vertex B. The center of the square M is the midpoint of the diagonal AC. Since A(2,7) and C(8,1), the center has coordinates
Point M is also the midpoint of the diagonal BD. Let B has coordinates (x,y), then
Hence, B(8,7).
Now, the dilation by a scale factor 2 with the center of dilation at the origin has the rule
(x,y)→(2x,2y).
Thus,
B(8,7)→B'(16,14).
Answer:
A is correct hope this helps
second answer to the left
The equation of the parabolas given will be found as follows:
a] general form of the parabolas is:
y=k(ax^2+bx+c)
taking to points form the first graph say (2,-2) (3,2), thus
y=k(x-2)(x-3)
y=k(x^2-5x+6)
taking another point (-1,5)
5=k((-1)^2-5(-1)+6)
5=k(1+5+6)
5=12k
k=5/12
thus the equation will be:
y=5/12(x^2-5x+6)
b] Using the vertex form of the quadratic equations:
y=a(x-h)^2+k
where (h,k) is the vertex
from the graph, the vertex is hence: (-2,1)
thus the equation will be:
y=a(x+2)^2+1
taking the point say (0,3) and solving for a
3=a(0+2)^2+1
3=4a+1
a=1/2
hence the equation will be:
y=1/2(x+2)^2+1
