Question 1014984
The path of the golf ball can be modeled by the quadratic equation
{{{y=- gx^2/20736 +x}}} , where
{{{x}}} is the ball's horizontal position (in feet),
{{{y}}} is its corresponding height (in feet), and
{{{g}}} is the acceleration due to gravity (in {{{feet/second^2}}} ).
 
On the moon, {{{g=5.32}}} , so the ball is an the moon's surface, at {{{y=0}}} when
{{{0=-5.32x^2/20736+x}}}-->{{{0=(-5.32/20736x-1)x}}}-->{{{system(x=0,"or",-5.32/20736x=1)}}}-->{{{system(x=0,"or",x=20736/5.32=highlight(about3898))}}}
There are two moments when the ball is on the moon's surface:
{{{x=0}}} (when the ball is hit and has not traveled horizontally), and
{{{x=highlight(about3898)}}} , when the ball hits the moon's surface after travelling a horizontal distance of about {{{highlight(3898feet)}}}(rounded).
 
When the golf ball is hit on Mars, where {{{g=12.17}}}{{{feet/second^2}}} ,
the path of the golf ball can be modeled by the quadratic equation
{{{y=-12.17x^2/20736 +x}}} , where
{{{x}}} is the ball's horizontal position (in feet), and
{{{y}}} is its corresponding height (in feet).
That equation represents a parabola
Your teacher wants you to remember that a quadratic function {{{y=ax^2+bx+c}}}
with {{{a<0}}}
represents a parabola with a vertex/maximum at
{{{x=-b/2a}}} ,
so you could apply that "formula" with {{{a=-12.17/20736}}} and {{{b=1}}} ,
and then you could substitute the {{{x=about852}}} value found into {{{y=-12.17x^2/20736 +x}}} to find the maximum height,
{{{y=-12.17*852^2/20736+852=about426}}}.
Alternately, you could "complete the square":
{{{y=-12.17x^2/20736 +x}}}
{{{(-20736/12.17)y=x^2+(-20736/12.17)x}}}
{{{(-20736/12.17)y+(20736/(2*12.17))^2=x^2+(-20736/12.17)x+(20736/(2*12.17))^2}}}
{{{(-20736/12.17)y+20736^2/(4*12.17^2)=(x-20736/(2*12.17))^2}}}
{{{(-20736/12.17)(y-20736/(4*12.17))=(x-20736/(2*12.17))^2}}}
So, {{{(-20736/12.17)(y-20736/(4*12.17))>=0}}}-->{{{y-20736/(4*12.17)<=0}}}-->{{{y<=20736/(4*12.17)=about426}}}} .
So, in Mars, the ball reaches a height of {{{highlight(426feet)}}} .