The archway of the main entrance of a university is modeled by the quadratic equation y = -x2 + 6x. The university is hanging a
banner at the main entrance at an angle defined by the equation 4y = 21 − x. At what points should the banner be attached to the archway?
1 answer:
Answer:
The point at which the banner should be attached to the archway is going to be given by the point(s) at which the two equations intercept:
y = -x^2 + 6x and 4y = 21 − x
Solving the system of equations, we find that they intercept at: (1, 5) and (5.25, 3.938)
Therefore, the banner should be attached to the archway at the points (1, 5) and (5.25, 3.938)
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Answer:
The system has one solution.
Both lines have the same y-intercept.
The solution is the intersection of the 2 lines.
Step-by-step explanation:
Lines are
slope = 0.5
y intercept(put x =0) =5
slope = 1
y intercept(put x =0 ) =5
Both are linear equation thus they have only one solution. Also two non parallel lines meet only at one point.
If you solve these linear equation coordinate of point of intersection of line on graph will come.