Parabola Equations - Algebra
Please explain working clearly.
Help greatly appreciated,
Thanks!

Answer:

[tex]\displaystyle h=-\frac{1}{6}x^2+2x[/tex]

Step-by-step explanation:

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

[tex]h=rx^2+tx[/tex]

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:

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

In our case, we have the function:

[tex]h=(r)x^2+(t)x[/tex]

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

Therefore, the x-coordinate of our vertex is:

[tex]\displaystyle x = -\frac{t}{2r}[/tex]

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

[tex]\displaystyle (6)=r\left(-\frac{t}{2r}\right)^2+t\left(-\frac{t}{2r}\right)[/tex]

Simplify:

[tex]\displaystyle 6=r\left(\frac{t^2}{4r^2}\right)-\frac{t^2}{2r}[/tex]

Simplify:

[tex]\displaystyle 6=\frac{t^2}{4r}-\frac{t^2}{2r}[/tex]

Subtract:

[tex]\displaystyle 6=\frac{t^2}{4r}-\frac{2t^2}{4r} = -\frac{t^2}{4r}[/tex]

Multiply:

[tex]\displaystyle 24r=-t^2, \, r\neq 0[/tex]

Solve for r:

[tex]\displaystyle r=-\frac{t^2}{24}[/tex]

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:

[tex]0=rx^2+tx[/tex]

Factor:

[tex]0=x(rx+t)[/tex]

By the Zero Product Property:

[tex]x=0\text{ or } rx+t=0[/tex]

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:

[tex]rx=-t[/tex]

Divide both sides by r:

[tex]\displaystyle x=-\frac{t}{r}[/tex]

We know that this zero must be 12. Thus:

[tex]\displaystyle 12=-\frac{t}{r}[/tex]

We have previously solved for r. Substitute:

[tex]\displaystyle 12=-\frac{t}{-\dfrac{t^2}{24}}[/tex]

Simplify:

[tex]\displaystyle 12=\frac{24}{t}[/tex]

Take the reciprocal of both sides:

[tex]\displaystyle \frac{t}{24}=\frac{1}{12}[/tex]

Multiply. Hence, the value of t is:

[tex]t=2[/tex]

Find r. Recall that:

[tex]\displaystyle r=-\frac{t^2}{24}[/tex]

Therefore:

[tex]\displaystyle r=-\frac{(2)^2}{24}=-\frac{4}{24}=-\frac{1}{6}[/tex]

Then in conclusion, our equation will be:

[tex]\displaystyle h=-\frac{1}{6}x^2+2x[/tex]

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