Can someone please help me with this function question on my lab? It's question #3. The lab is attached

Question

anonymous · Answer

attachment

campbell_st · Answer

quick question.... have you done calculus

anonymous · Answer

No

anonymous · Answer

The class is math 111 college algebra. I haven't seen a problem like this, not sure where to start.

campbell_st · Answer

ok... so you'll need to find the line of symmetry.... that max height will lie on the line... 

so for a quadratic 

$$ac^2 +bx + c = 0$$

the line of symmetry is 

$$x = \frac{-b}{2a}$$

in your question b = 2

and   a = -0..05

substitute to find the x value that givens max height... 

then substitute it into the equation to find how high the object goes

hope that helps

campbell_st · Answer

for part (b) subsitute = 390

and then find the height.... if the answer is greater than 20... then it clears the fence.

campbell_st · Answer

for (a) 

$$x = \frac{-2}{2 \times -0.005}$$

that gives the value of x when the max height occurs... substitute it into the the equation to find the max height

for (b)

you need 

$$h(390) = -0.005 \times (390)^2 + 2 \times 390 + 3.5$$

anonymous · Answer

203.5 for part a? 23 for part b?

campbell_st · Answer

(a) is great..

(b) does it clear the fence....?

(c) you need to solve for x, I'd recommend the general quadratic formula to find how far it travels...

(d) similar to (b) substitute x = 375... 

hope it all helps

anonymous · Answer

B it does clear the fence? D= 751.625?

anonymous · Answer

my bad 50.375 for D

campbell_st · Answer

well done

anonymous · Answer

Thanks I really appreciate your help!