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 = ­0.005x2 + 0.45x + 125. will the temperature of the part ever reach or exceed 132 ° f? use the discriminant of a quadratic equation to decide.

.005x^2+.45x+125=132

.005x^2+.45x-7=0

So the discriminant, the (b^2-4ac) from the quadratic equation is:

0.3425

Since the discriminant is positive, there are two real solutions, this implies that the temperature will reach 132 at two times. However

x=(-.45±√.3425)/(.01)

x≈-103.52, 13.52

As you can see only the positive solution has any meaning as there is no negative running time for the machine.  And since the derivative and second derivative are positive for this function, the temperature keeps increasing after the machine reaches 132F.  So the temperature reaches and exceeds 132F for all times greater than or equal to 13.52 minutes.

the formula of a slope:

from a table we have

substitute:

the slope of f(x) is -1.

the graph of funcfion g(x) is a horizontal line. the slope of a horizontal line is equal 0. therefore the slope of g(x) is 0.

-1 < 0

answer: the function g(x) has the greatest slope.

Ithink it's b sorry if i'm wrong