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Sagot :
we have that
[tex]y=-x^2+36[/tex]
we know that
Is the equation of a vertical parabola open down
so
the vertex is a maximum
step 1
convert the equation of the parabola in the vertex form
[tex]y=-x^2+36[/tex]
[tex]y-36=-x^2[/tex]
the vertex (h,k) is the point (0,36)
Part a) The point on the graph where the height of the tunnel is a maximum is (0,36)
Part b) The points on the graph where the height of the tunnel is zero feet is when y=0
so
for y=0
[tex]-x^2+36=0[/tex]
[tex]x^2=36[/tex]
[tex]x=(+/-)6[/tex]
the points are (-6,0) and (6,0)
see the attached figure

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