Try this solution with a short explanation:
a) y=ax²+5; ⇒ substitution (4;0): 16a+5=0; ⇒ a= -5/16.
Answer: <u>y= -5/16 x²+5</u>;
b)<u> y=x²</u>;
c) y=ax(x+7); ⇒ substitution (4;4): 4a(4+7)=4; ⇒a=1/11.
Answer: <u>y= 1/11 x²+ 7/11 x</u>;
d) y=a(x-1)(x-3); ⇒substitution (0;3): (-1)*(-3)*a=3; ⇒a=1.
Answer: <u>y=x²-4x+3</u>;
e) y=a(x-1)(x+3); ⇒ substitution (-1;5): (-2)*2*a=5; ⇒ a= -5/4.
Answer: <u>y= -5/4 x² -5/2 x +15/4</u>;
f) y=ax²+bx+c; ⇒ substitution (2;2): 4a+2b+6=2 or 2a+b= -2 (the I-st equation of the system);
the x-coordinate of the vertex: -b/2a=2; ⇒ b= -4a (the II-d equation of the system);
Answer: <u>y=x²-4x+6.</u>