Question

The arch of a train tunnel is in the shape of a parabola. The maximum height of the tunnel is 20 m. If the origin of a grid is placed at one end of the arch where the tunnel contacts the road, the other point of contact with the road is 40 m away. Which quadratic function can be used to model the arch?

Answer 1

` If the origin of a grid is placed at one end of the arch ` -- so that would be the position of the parabola ; i am attaching an image of it

we can see that he roots of it will be \(( 0,0) \) and \( (40,0) \)

its vertex is going to be \( (20,20) \)



the quadratic will be in the form of \(y = ax^2+ bx +c \) where in out case, \(c\) is \(0 \)

the axis of symmetry is \(x=20 \) right? cuz its in the middle of it ; \( y\) is going to bemax height, which is \( 20 \) -- got that from the vertex

so we do:
\[ 20 = a\times 20^2+ b \times 20 \cancel{+0 } \]
\[ 20=400a+20b \]
now th eroots, two one them -- the one which is at \( (0,0) \) -- logically we are going to end up with \( a=0 \) and \( b=0 \) which indicate absolutely nothing, so forget about that one root

as for the second one, which is at \( (40,0) \) we do the thing again:
\[ 0 = a\times 40^2+ b \times 40 \cancel{+0 } \]
\[ 0=1600a+40b \]
now we have to do whith this system:
\[ \begin{cases} 20=400a+20b \\ 0=1600a+40b \end{cases} \]

solve the simple system, find the value of b and a, and plug that in the main formula

Answer 2

\[ \begin{cases} 20=400a+20b \\ 0=1600a+40b \end{cases} \]

ok so i will do it by substituion -- you can chose whichever way you want to do it
\[ 20=400a+20b \Rightarrow 20-20b=400a \Rightarrow \frac{ 20-20b}{400}=a \]
\[ \frac{ \cancel{20}-\cancel{20}b}{\cancel{400}}=a \Rightarrow \frac{1-b}{20} =a \]
lets plug that into the 2nd equation:
\[ 0=1600a+40b \Rightarrow 0=1600\times \frac{1-b}{20} +40b \Rightarrow 0=\frac{(1-b)\times 1600}{20} +40b \]
\[ \ \ \ \ \ \ \ \ \Updownarrow \]
\[ 0=80(-b+1)+40b \Rightarrow 0=-80b+80+40b \]
\[ 0=-40b+80 \Rightarrow b=2 \]
as mentioned earlier, \( a= \frac{1-b}{20} \) , or just \( a= -\frac{1}{20} \)

thus, the equation of that parabola would be:
\[ y = ax^2+ bx +c \; \; \; \;====\Rightarrow \;\;\;\;\;\; y = -\frac{1}{20} x^2+2x+0 \]
and that would be the answer :)

Answer 3

Im still confused. I understand the beginning but don't understand the end