Find the coordinates of the point of intersection of all parabolas that have equation y=x2+px +q with p+q=2020. The point of intersection is​

(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

the mistake in the pattern is that :

b. the decimal point was moved one too many place values each time

the correct pattern could be made by :

b. moving the decimal point to the right the same numbers of place values as the exponents