Find the equations of the lines...
First find the slope...(y2-y1)/(x2-x1)=m
m=(6-1)/(8-5)=5/3 and it passes through (5,1) so
y=mx+b becomes:
y=5x/3+b and using (5,1)
1=5(5)/3+b
1=25/3+b
3/3-25/3=b
b=-22/3 so
y1=(5x-22)/3
.... now y2...
m=(8-3)/(-1--4)=5/3 (note it has the same slope as y1...
y=5x/3+b and using the point (-1,8)
8=5(-1)/3+b
8=-5/3+b
24/3+5/3=b
b=29/3, now note that the y-intercept is different...
y2=(5x+29)/3
Since these lines have the same slope but different y-intercepts, they are parallel to each other. (and will never intersect.)
I hope this helps you
Area=6^2pi=36pi
Circumference =2.6pi=12pi
Answer:
We can have two cases.
A quadratic function where the leading coefficient is larger than zero, in this case the arms of the graph will open up, and it will continue forever, so the maximum in this case is infinite.
A quadratic function where the leading coefficient is negative. In this case the arms of the graph will open down, then the maximum of the quadratic function coincides with the vertex of the function.
Where for a generic function:
y(x) = a*x^2 + b*x + c
The vertex is at:
x = -a/2b
and the maximum value is:
y(-a/2b)
