Question

Parabola Equations - Algebra

Answer 1

Answer:

$h=-\frac{1}{6}x^2+2x$

Step-by-step explanation:

The height of the tunnel is modeled by:

$h=rx^2+tx$

Where r and t are constants.

We know that the maximum height of the tunnel h is 6 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.

The vertex is given by:

$(-\frac{b}{2a}, f(-\frac{b}{2a}))$

In our case, we have the function:

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

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

Therefore, our vertex is:

$\Rightarrow -\frac{t}{2r}$

Thus, if we substitute this back into our equation, we should get 6 since 6 is the maximum height which is determine by the vertex. In other words:

$(6)=r(-\frac{t}{2r})^2+t(-\frac{t}{2r})$

Simplify:

$6=r(\frac{t^2}{4r^2})-\frac{t^2}{2r}$

Simplify:

$6=\frac{t^2}{4r}-\frac{t^2}{2r}$

Combine fractions:

$6=\frac{t^2}{4r}-\frac{2t^2}{4r}=\frac{-t^2}{4r}$

Multiply both sides by the denominator:

$24r=-t^2$

Let's solve for the constant r. Divide both sides by 24. Hence:

$r=-\frac{t^2}{24}$

Now, we can use our knowledge of the width.

We know that the width is 12 meters at ground level.

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

So:

$0=rx^2+tx$

We can factor:

$0=x(rx+t)$

By the Zero Product Property:

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

So, 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:

$x=-\frac{t}{r}$

We know that this zero must be 12. Thus:

$12=-\frac{t}{r}$

We have previously solved for r. Substitute:

$12=-\frac{t}{\frac{-t^2}{24}}$

Simplify:

$12=\frac{24}{t}$

Take the reciprocal of both sides:

$\frac{t}{24}=\frac{1}{12}$

Multiply both sides by 24. Hence, the value of t is:

$t=2$

Now, we can find r. r is given by:

$r=-\frac{t^2}{24}$

By substituting 2 for t, then, we acquire that:

$r=-\frac{(2)^2}{24}=-\frac{4}{24}=-\frac{1}{6}$

Therefore, for:

$h=rx^2+tx$

Our equation is:

$h=-\frac{1}{6}x^2+2x$