Question

I NEED HELP PRE-CAL i only have hours left!

Answer 1

If you take the point on the bridge directly beneath the lowest point on the cable to be the origin, then the parabola has equation

$y = ax^2+bx+10$

300 ft to either side of the origin, the parabola reaches a value of y = 90, so

$a(300)^2+b(300)+10 = 90 \implies 4500a+15b = 4$

$a(-300)^2+b(-300)+10 = 90 \implies 4500a-15b = 4$

Adding these together eliminates b and lets you solve for a :

$(4500a+15b)+(4500a-15b)=4+4 \\\\ 9000a = 8 \\\\ a = \dfrac1{1125}$

Solving for b gives

$4500\left(\dfrac1{1125}\right)+15b=4 \\\\ 4+15b = 4 \\\\ 15b=0 \\\\ b=0$

So the parabola's equation is

$y = \dfrac{x^2}{1125}+10$

150 ft away from the origin, the cable is at a height y of

$y = \dfrac{150^2}{1125}+10 = \boxed{30}$

ft above the bridge.