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 135 degrees. the temperature T in degrees fahrenheit x minutes after the machine is put into operation is modeled by T=-0.005x^2+0.45x+123

a. tell whether the temperature of the part will ever reach or exceed 135 degrees Fahrenheit. use the discriminant of a quadratic equation to decide.

b. if the machine is in operation for 90 minutes before being turned off, how many times will the temperature of the part be 154 degrees Fahrenheit?

As x gets very big positive or negative, T gets very small. In other words this parabola opens down (sheds water)

Where is the vertex (the top) ?
Complete the square
.005 x^2 - .45 x -123 = -T

x^2 - 90 x - 24600 = -200 T

x^2 - 90 x = -200 T +24600

x^2 -90 x + 2025 = -200 T + 26625

(x-45)^2 = -200 (T - 133.125)
so
vertex at (45 , 133.125)
so
It does not quite make 135 degrees

Now wait a minute, it never reaches 154 degrees.

To answer these questions, we need to analyze the given quadratic equation and apply the appropriate principles. Let's break down each question and explain the steps to solve them:

a. To determine if the temperature of the part will ever reach or exceed 135 degrees Fahrenheit, we need to check if the quadratic equation has any real solutions above 135. The discriminant of a quadratic equation, denoted by Δ (delta), is the expression inside the square root of the quadratic formula (b^2 - 4ac).

In our equation T = -0.005x^2 + 0.45x + 123, the coefficients are: a = -0.005, b = 0.45, and c = 123. The discriminant can be calculated as follows:

Δ = b^2 - 4ac

Plugging in the values, we have:

Δ = (0.45)^2 - 4(-0.005)(123)

Simplifying further:

Δ = 0.2025 + 2.46

Δ = 2.6625

Since Δ is positive (greater than zero), the quadratic equation has two distinct real solutions. Therefore, the temperature of the part will reach or exceed 135 degrees Fahrenheit at some point during the manufacturing process.

b. To find out how many times the temperature of the part will be 154 degrees Fahrenheit when the machine is in operation for 90 minutes, we need to substitute the value of x = 90 into the quadratic equation and solve for T.

T = -0.005x^2 + 0.45x + 123

Substituting x = 90:

T = -0.005(90)^2 + 0.45(90) + 123

Calculating further:

T = -0.005(8100) + 40.5 + 123

T = -40.5 + 40.5 + 123

T = 123

Thus, when the machine is in operation for 90 minutes, the temperature of the part will be 154 degrees Fahrenheit only once, as given by T = 123.