Please help me with my quadratic function word problem!

Question

The height y of a parabolic arch is given by y=-1/16x²+40, where x is the horizontal distance from the center of the base of The arch. All distances are in feet. What is the highest point on the arch? How wide is the arch at the base to the nearest tenth of a foot?

Answer 1

Ok, Riley...

If I don't finish is because the Library is about to close, and I type slowly.

 

First, of all, the horizontal distance from the center of the arc will be 40 ft.  The +40 means that you are moving the parabola up 40 from the origin, (0,0).  Just move it straight up.  But remember since there is a neg. 1/16 in front of X^2, the parabola is concave down, a u upside down.  So, the highest point, will be 40 feet above the ground.  Now, to find how wide the arch is, we need to find the x intercepts.  So, make y= 0.  So, we have

 

0= (-1/16)x^2+40        Subtract 40from both sides

-40                  -40

-40= (-1/16)x^2          Multiply by -16 on both sides

640= x^2

=/- 25.3 ft= x 

 

Double that amount, 50.6 is how wide the arch will be.  Do to the internet and draw the parabola, it is easier to explain it when you actually see the figure.

 

Ok, Riley.. I hope this helps.. Go to sleep.. and please rate my answer.. take care..

 

D. Y. Taylor