Question

During a manufacturing process, a metal part in a machine is exposed to varying temperature conditions. The manufacturer of the machine recommends that the temperature of the machine part remain below 132 ° F. The temperature T in degrees Fahrenheit x minutes after the machine is put into operation is modeled by T = -0.005x^2 + 0.45x + 125. Will the temperature of the part ever reach or exceed 132 ° F? Use the discriminant of a quadratic equation to decide.

Answer 1

We are going to solve this problem two ways:

Solution 1. Graphical

If you plot the equation of temperature $y(x) = -0.005x^2 + 0.45x + 125$ for x between 0 and 160 minutes you'll get the plot in the attachement.

The blue curve represents the variation of the temperature in time. Notice that it crosses the red line (132 degrees Fahrenheit) twice, at around 20 F and 70 F.

So the answer is yes, if will reach 132 F.

Solution 2. Analytical

In order to see if the variation of temperature y(x) intersects 132 F line means to solve the equation:

$-0.005x^2 + 0.45x + 125 = 132$

Step 1. Write the equation in the quadratic form:
$-0.005x^2 + 0.45x - 7 = 0$

Step 2. Calculate the discriminant:
$\Delta = b^2 - 4ac = 0.45^2-4 \cdot (-0.005) \cdot (-7) = 0.06$

Only by looking at the discriminant we can see that it is positive, which means that it crosses the 132 F line in two points, which also means that there are two roots of the equation, both real numbers.

Step 3 (optional). Calculate the roots of the equation.

$x_1 = \frac{-b+ \sqrt{\Delta}}{2a} = \frac{-0.45+ \sqrt{0.06}}{2 \cdot (-0.005)}=20.5$
$x_2 = \frac{-b- \sqrt{\Delta}}{2a} = \frac{-0.45- \sqrt{0.06}}{2 \cdot (-0.005)}=69.5$

As you can see, tha analytic solution matches the graphic solution. At time x=20.5 minutes the temperatures reaches 132 F and is still rising. At time x=69.5 minutes the temperature is again 132 F but this time is decreasing.