A tunnel with a parabolic arch is 12m wide. If the height of the arch 4m from the left edge is 6m, can a truck that is 5m tall and 3.5 m wide pass through the tunnel? HELP!
I kind of have the answer but I don't know what to do next..
yes
but why?
Think about it if 4m away it its 6m tall then 6m away its going to grow. Every two meters twords the middle it grows three meters up
y = a(x-r)(x-s) = a (x-12)(x-0) 6 = a(4-12)(4-0) 6 = a(-8)(4) 6 = -32a -- ---- -32 -32 6 = a - -- 32 - 3 -- = a 16
but I need to prove my justification using quadratic equations.
Ok so you know it can fit width wise. So we need to find height now
2nd: y = -3/16 x(x-12) 5 = -3/16x(x-12) 16*5 = -3/16*16x(x-12) 80 = -3x^2 + 36x 3x^2 -36x + 80 = 0 a = 3, b = -36, c = 80 x = -(-36) +/- √(-36)^2 - 4(3)(80) --------------------------- 2(3)
= 36 +/- √336 ------------ 6 x = 36 + √336 ----------- 6 = 9.1 or x= 36 - √336 ---------- 6 = 2.9
that kinda makes sense. o me this is not a well written question considering its given the car can fit but Ill try and help
oh, thank you Jacob. & the answer key says: f(x) = -3/16x (x-12) Yes, because at a height of 5 m the bridge is 6.11 m wide. How is it 6.11 m wide? o.o
I forgot what my teacher told me eep >_<"
Oh ok. Think of this as a parabola for the y value of 5 the x value is 6.11. Th bridge up to that point is 6.11m wide
Does that kinda make sense?
no not really.. D:
ohhh, wait yes I get that the x value will be 6.11 but how? o.o
try plugging 5 as the y value to one of your equatons
I tried adding the zeros and divide by 2 but it's 10.31 not 6.11 o.o
What you need to do is find the equation of the parabola with some of the given points
then plug in 5
That's what I did in the beginning and I'm still confused as to how they got 6.11...
I'm sorry. I dont know the answer I can ask my teacher tommorow if you want he can help.
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