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Parabola Equations - Algebra

Please explain working clearly.
Help greatly appreciated,
Thanks!

Parabola Equations - Algebra Please explain working clearly. Help greatly appreciated-example-1

1 Answer

7 votes

Answer:


\displaystyle h=-(1)/(6)x^2+2x

Explanation:

We are given that the height of the tunnel is modeled by:


h=rx^2+tx

Where r and t are constants.

We are also given that the maximum height of the tunnel h is six meters.

And at ground level, the width is 12 meters.

And we want to determine the equation of the parabola.

First, since this is a quadratic, our maximum height h will occur at the vertex of our equation.

Recall that the vertex is given by the formulas:


\displaystyle \text{Vertex} = \left(-(b)/(2a), f\left(-(b)/(2a)\right)\right)

In our case, we have the function:


h=(r)x^2+(t)x

Hence, a = r; b = t; and c = 0.

Therefore, the x-coordinate of our vertex is:


\displaystyle x = -(t)/(2r)

Therefore, if we substitute this back into our equation, the result should be six since six is the maximum height. Hence:


\displaystyle (6)=r\left(-(t)/(2r)\right)^2+t\left(-(t)/(2r)\right)

Simplify:


\displaystyle 6=r\left((t^2)/(4r^2)\right)-(t^2)/(2r)

Simplify:


\displaystyle 6=(t^2)/(4r)-(t^2)/(2r)

Subtract:


\displaystyle 6=(t^2)/(4r)-(2t^2)/(4r) = -(t^2)/(4r)

Multiply:


\displaystyle 24r=-t^2, \, r\\eq 0

Solve for r:


\displaystyle r=-(t^2)/(24)

Next, we are given that the width is 12 meters at ground level.

Hence, when h = 0, the difference of our roots is 12:

We have:


0=rx^2+tx

Factor:


0=x(rx+t)

By the Zero Product Property:


x=0\text{ or } rx+t=0

Hence, the first zero is 0.

Therefore, the second zero must be 12 to ensure that our width is 12.

Let's isolate the second zero. Subtract t from both sides:


rx=-t

Divide both sides by r:


\displaystyle x=-(t)/(r)

We know that this zero must be 12. Thus:


\displaystyle 12=-(t)/(r)

We have previously solved for r. Substitute:


\displaystyle 12=-(t)/(-(t^2)/(24))

Simplify:


\displaystyle 12=(24)/(t)

Take the reciprocal of both sides:


\displaystyle (t)/(24)=(1)/(12)

Multiply. Hence, the value of t is:


t=2

Find r. Recall that:


\displaystyle r=-(t^2)/(24)

Therefore:


\displaystyle r=-((2)^2)/(24)=-(4)/(24)=-(1)/(6)

Then in conclusion, our equation will be:


\displaystyle h=-(1)/(6)x^2+2x

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