I will give brainliest just please help The equation ax2 + bx + c = 0 has no real solutions.
Which statement about the graph of f(x) = ax2 + bx + c could be true?

A. It could pass through the origin.
B. Its vertex could be at (–6, 0).
C. It could have a maximum at (–3, 2).
D. It could have a minimum at (0, 4).

D. It could have a minimum at (0, 4).

Step-by-step explanation:

ax2 + bx + c = 0  has no real solutions, which means it does not touch the x axis

Since it points upwards, the vertex must be above the x axis

It will have a minimum since it opens upwards

A. It could pass through the origin.   False

B. Its vertex could be at (–6, 0).   False

C. It could have a maximum at (–3, 2).   False

D. It could have a minimum at (0, 4).  True

answer: the answer is ann

< xcy ≅ < xby

step-by-step explanation:

in δcax and δbax,

so, δcax ≅ δbax (side-angle-side or sas)

since δcax ≅ δbax, we can conclude < acx ≅ < abx

in δacy and δaby,

so, δacy ≅ δaby (side-angle-side or sas)

since δacy ≅ δaby, we can conclude < acy ≅ < aby

now,

< acy ≅ < aby

=> < acx + < xcy ≅ < abx + < xby

=> < acx + < xcy ≅ < acx + < xby   (since < acx ≅ < abx already proved above)

subtracting < acx from both sides, we get

< acx + < xcy -< acx ≅ < acx + < xby -< acx

cancelling out < acx and -< acx from both sides, we get

< xcy ≅ < xby (proved)