A football is kicked at 40 yards away from a goal post that is 10 feet high. Its path is modeled by y = -0.03x^2+1.6x, where x is the horizontal distance in yards traveled by the football and y is the corresponding height above the ground in feet.

Does the football go over the goal post? How far above or below the goal post is the football? Use two or more complete sentences to explain your answers. (Hint: The unit conversion is built into the given function.)

Question

A football is kicked at 40 yards away from a goal post that is 10 feet high. Its path is modeled by y = -0.03x^2+1.6x, where x is the horizontal distance in yards traveled by the football and y is the corresponding height above the ground in feet.

Does the football go over the goal post? How far above or below the goal post is the football? Use two or more complete sentences to explain your answers. (Hint: The unit conversion is built into the given function.)

ellie202000 · Answer

I tried to plug in 40 for x but I am pretty sure that's wrong.

campbell_st · Answer

well you have the correct method.... if you plug in the horizontal distance of x = 40 you will get the height..y.... if y > 10 the ball goes over the goat...

campbell_st · Answer

oops goal and not goat

ellie202000 · Answer

ok so the formula would look like$$10=-0.03(40)^{2}+1.6(40)$$

campbell_st · Answer

basically yes... I would have said

f(40) = .....

campbell_st · Answer

then decide if f(40) > 10   goes over   f(40) < 10 goes under f(40) = 0 hits the bar and who knows...

campbell_st · Answer

oops hits the bar should be f(40) = 10

ellie202000 · Answer

which would be $$f(40)=-0.03(40)^{2}+1.6(40)$$$$f(40)=48+64$$$$f(40)=112$$$$112>10$$ so it would go over? But 112 just seams to high of a number.

campbell_st · Answer

I think you need to check the calculation 

$$-0.03 \times 40^2 + 1.6 \times 40 = 16$$

ellie202000 · Answer

Oh I forgot the negative! That makes sense now. Thank you so much for all your help.