Question 1019411
Standard Form equation for a parabola, {{{y=a(x-h)^2+k}}}.


<i>Water issuing from the end of a horizontal pipe 30 ft. above the ground describes a parabolic curve, where the vertex is at the end of the pipe.</i>
{{{y=a(x-0)^2+30}}}


<i>At a point 10 ft. below the pipe, the water is 12 ft. from the vertical line through the end of the pipe. </i>
{{{30-10=a(12-0)^2+30}}}
{{{20=144a+30}}}
{{{144a=-10}}}
{{{a=-10/144}}}
{{{a=-5/72}}}


{{{highlight(highlight_green(y=-(5/72)x^2+30))}}}


The question means, find x for y=0.


A graph of this could be either the left branch or the right branch of the parabola, to correspond with the representation of the situation for the end of the hose being the vertex (a maximum point).  The equation is done in standard form, so easier to graph.
{{{graph(300,300,-30,30,-10,50,-(5/72)x^2+30)}}}