Let's consider the given function.

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

The

y -

variable describes the

height

in meters

of

the

arch

above

the water .

We want to find the values of

x

for which the arch of the bridge is above the road. From the picture, we can see that the bridge is

52

meters above the water. Therefore, we want to find values of

x

such that

y > 52

. Let's find them!

y > 52

Substitute

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

- 0.00211 x^{2} + 1.06 x > 52

SubIneq

$LHS - 52 > RHS - 52$

- 0.00211 x^{2} + 1.06 x - 52 > 0

We got an quadratic inequality. There are many ways to solve it, but we will solve it algebraically. To do this, we will follow three steps.

Solve the related quadratic equation.
Plot the solutions on a number line.
Test a value from each interval to see if it satisfies the original inequality.

Step $1$

We will start by solving the related equation.

- 0.00211 x^{2} + 1.06 x - 52 > 0 ⇕ - 0.00211 x^{2} + 1.06 x + (- 52) > 0

We see above that

a = - 0.00211,

b = 1.06,

and

c = - 52 .

Let's substitute these values into the Quadratic Formula to solve the equation.

x = \frac{- b \pm b ^{2} - 4 a c}{2 a}

SubstituteValues

Substitute values

x = \frac{- 1 . 0 6 \pm 1 . 0 6 ^{2} - 4 ( - 0 . 0 0 2 1 1 ) ( - 5 2 )}{2 ( - 0 . 0 0 2 1 1 )}

UseCalc

Use a calculator

x \approx \frac{- 1 . 0 6 \pm 0 . 8 2 7 4 8}{- 0 . 0 0 4 2 2}

Now we can calculate the first root using the positive sign and the second root using the negative sign.

$x \approx \frac{- 1 . 0 6 \pm 0 . 8 2 7 4 8}{- 0 . 0 0 4 2 2}$
$x \approx \frac{- 1 . 0 6 + 0 . 8 2 7 4 8}{- 0 . 0 0 4 2 2}$	$x \approx \frac{- 1 . 0 6 - 0 . 8 2 7 4 8}{- 0 . 0 0 4 2 2}$
$x \approx 55$	$x \approx 447$

Step $2$

The solutions of the related equation are approximately $55$ and $447 .$ Let's plot them on a number line. Since the original one is a strict inequality, the points will be open.

Step $3$

Finally, we must test a value from each interval to see if it satisfies the original inequality. Let's choose a value from the first interval,

x < 55 .

For simplicity, we will choose

x = 0 .

- 0.00211 x^{2} + 1.06 x - 52 > 0

Substitute

$x = 0$

- 0.00211 (0)^{2} + 1.06 (0) - 52 > 0

CalcPowProd

Calculate power and product

0 + 0 - 52 > 0

AddSubTerms

Add and subtract terms

- 52 ≯ 0 \times

Since

x = 0

produced a false statement, the interval

x < 55

is not part of the solution. Similarly, we can test the other two intervals.

Interval	Test Value	Statement	Is it part of the solution?
$55 < x < 447$	$100$	$32.9 > 0$ $✓$	Yes
$x > 447$	$500$	$- 49.5 ≯ 0$ $\times$	No

We can now write the solution set.

55 < x < 447

Since the

x -

variable represents the distance in meters from the left pylon, the arch is above the road about

55

m from the left pylon to about

447

m from the left pylon.

Step 1

Step 2

Step 3

Step $1$

Step $2$

Step $3$

Exercise