Question

The arch of the Sydney Harbor Bridge in Sydney, Australia, can be modeled by y=−0.00211x2+1.06x, where x is the distance (in meters) from the left pylons and y is the height (in meters) of the arch above the water. For what distances x is the arch above the road? Round your answers to the nearest whole number.

Answer 1

Answer:

The arch above the road is 251 meters from the left pylons

Step-by-step explanation:

we have

$y=-0.00211x^{2}+1.06x$

This is a vertical parabola open downward (because the leading coefficient is negative)

The vertex is a maximum

The y-coordinate of the vertex represent the maximum height of the arch above the road

Remember that the x-coordinate of the vertex is the midpoint between the roots of the quadratic equation

One root is the origin (0,0)

Determine the second root

For y=0

$-0.00211x^{2}+1.06x=0$

$0.00211x^{2}=1.06x$

Simplify

$0.00211x=1.06$

$x=502$

Find the midpoint between the roots

$x_m=(0+502)/2=251$

so

The x-coordinate of the vertex is 251

therefore

The arch above the road is 251 meters from the left pylons

see the attached figure to better understand the problem

Answer 2

Answer:

The arch is above the road for a horizontal distance of 502.37 meters.

Step-by-step explanation:

The Arch is modeled by

$y=-0.00211x^{2} +1.06x$

Which represents a parabola. Remember that quadratic equations represent parabolas.

Assuming that the road is the x-axis, we can get the answer by using $y=0$, and solving the equation for $x$

$0=-0.00211x^{2} +1.06x\\x(-0.00211x+1.06)=0$

Now, we use the zero property

$x_{1} =0\\-0.00211x_{2}+1.06=0 \implies x_{2}=\frac{-1.06}{-0.00211} \approx 502.37$

Therefore, the arch is above the road for a horizontal distance of 502.37 meters.