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:
y=1/2x+2
Step-by-step explanation:
The equations come in the form of y=mx+c, where m is the gradient and c is the y-intercept. Looking at the graph we know that the y-intercept is (0,2), so that rules out options B and C.
To find the gradient is a little more tricky, but we can follow the formula:
, where rise is the vertical value and run is the horizontal value
So we sub in our values:
And simplify:
So now we sub in our new found values into y=mx+c:
y=1/2x+2