Answer:
(1,2021)
Step-by-step explanation:
P and q can vary subject to their sum being 2020.
Consider one parabola with p1 and q1 and another with p2 and q2.
y1=(x1)^2+(p1)(x1)+(q1)
y1=(x2)^2+(p2)(x2)+(q2)
At their intersection, the x and y coordinates are the same.
y1=y2=y
x1=x2=x
x^2+(p1)x+(q1)=x^2+(p2)x+(q2)
Solve for x
x(p1-p2)=q2-q1
x=(q2-q1)/(p1-p2)
Use the constraint that p+q=2020 to eliminate p1 and p2.
p1=2020-q1
p2=2020-q2
x=(q2-q1)/(2020-q1-2020+q2)
x=(q2-q1)/(q2-q1)
x=1
Substitute in the equation for y.
y=1^2+p(1)+q
y=2021
F(3)=53 and for f(x)=1; x=2
Answer:
(x,y)= (5/13 , 55/13)
Step-by-step explanation:
Answer:
Its A
Thats because if you draw it then you will find that the vertex is 8, -8
A,
(8, -8)
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