Question 1120311
the equation is T = -0.005 * x^2 + .45 * x + 125.


the equation is in standard form of ax^2 + bx + c


a = -.005
b = .45
c = 125


the maximum value of T is when x = -b/2a.

 
that would be when x = -.45 / (2*-.005)


simplify to get x = -.45 / -.01


solve for x to get x = 45


when x = 45, T = -.005 * x^2 + .45 * x + 125 = 135.125 degrees.


yes, the part will reach or exceed 135 degrees.


if the part is in operation for 90 minutes, the part will have reached 134 degrees two times.


you can replace T with y and graph the equation of y = -.005 * x^2 + .45 * x + 125.


the graph looks like this:


<img src = "http://theo.x10hosting.com/2018/072304.jpg" alt="$$$" >


from the graph, you can see that the part will reach or exceeed 134 degrees when x = 30 through x = 60.


that's a period of 30 minutes.


from the graph, you can also see that the maximum value of the temperature is 135 degrees.


the part will be 125 degrees when x = 0 and when x = 90, according to the formula.


the points at which the temperature reaches 134 degrees was found by taking the equation of y = -.005 * x^2 + .45 * x + 125 and replacing y with 134.


you get 134 = -.005 * x^2 + .45 * x + 125.


subtract 134 from both sides of this equation to get -.005 * x^2 + .45 * x - 9 = 0


factor this equation to get x = 30 or x = 60.


i used an online quadratic equation solver at <a href = "https://www.mathsisfun.com/quadratic-equation-solver.html" target = "_blank">https://www.mathsisfun.com/quadratic-equation-solver.html</a>


the results from that quadratic equation solver are shown below:


<img src = "http://theo.x10hosting.com/2018/072303.jpg" alt="$$$" >


if the quadratic equation solver is unavailable, you would solve using the quadratic formula.


that formula is referenced below:


<a href = "http://www.purplemath.com/modules/quadform.htm" target = "_blank">http://www.purplemath.com/modules/quadform.htm</a>