Yes, you can add a negative and a positive when doing like terms.
A, C, and E.
you can check them by putting into y=mx+b form. if m is negative then as domain increases range decreases.
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
Answer:
For Lin's answer
Step-by-step explanation:
When you have a triangle, you can flip it along a side and join that side with the original triangle, so in this case the triangle has been flipped along the longest side and that longest side is now common in both triangles. Now since these are the same triangle the area remains the same.
Now the two triangles form a quadrilateral, which we can prove is a parallelogram by finding out that the opposite sides of the parallelogram are equal since the two triangles are the same(congruent), and they are also parallel as the alternate interior angles of quadrilateral are the same. So the quadrilaral is a paralllelogram, therefore the area of a parallelogram is bh which id 7 * 4 = 7*2=28 sq units.
Since we already established that the triangles in the parallelogram are the same, therefore their areas are also the same, and that the area of the parallelogram is 28 sq units, we can say that A(Q)+A(Q)=28 sq units, therefore 2A(Q)=28 sq units, therefore A(Q)=14 sq units, where A(Q), is the area of triangle Q.
