a.
The red hose follows the path y=-(x-3)^2+7.
y=-(x-3)^2+7 is the parent function y=x^2:
shifted 3 units to the right, reflected with respect to the x-axis and shifted 7 units up.
So the vertex of the parent function becomes (3, 7)
Thus, the green hose throws the water higher.
b.Check the picture attached. Consider the green hose.
Let the nozzle of the hose be at (0, 0) and the top of the flower be at (10, 0).
The path of the water is described by the parabola with x-intercepts 0 and 10,
and vertex (5, 4).
Remark: the x-axis is not the ground level, but 4 feet high, this is why the vertex is (5, 8-4)=(5, 4).
the equation of a parabola with x-intercepts x=0 and x=10 is:
y=ax(x-10), to find a, we use the third point we have, the vertex:
y=ax(x-10)
4=a*5(5-10)
4=a*5(-5)
a=-4/25
Thus the parabolic path of the water from the green rose is described by:
y=-4/25x(x-10).
To avoid confusion in our final answer, we shift the parabola 4 units up, or better say, we shift the x-axis 4 units down: y=-4/25x(x-10)+4
c.
consider the value of x for which y=-4, which means the point x for which the water reaches the ground level.
-4=-4/25x(x-10)
1=1/25x(x-10)
25=x(x-10)
[tex]x^2-10x-25=0\\\\x^2-10x+25-50=0\\\\(x-5)^2-50=0\\\\x-5=\sqrt{50}=\sqrt{2*25}=5\sqrt{2}\\\\\x=5+5\sqrt{2}[/tex]
remark: we did not consider the case [tex]x-5=-\sqrt{50}[/tex], which would give us a negative x.
Domain: [tex][5+5\sqrt{2} ][/tex]
Answer:
a: the green hose
b: y=-4/25x(x-10)+4
c: domain of the green hose: [tex][5+5\sqrt{2} ][/tex]
