Question 102
<pre><font face = "courier new">1. let f be a quadratic function with a minimum value of
-22100, a line of symmetry of x=140, and a y-intercept of
(0,-2500).
2. the function g is the same as f but relfected about the
x-axis and reduced by a factor of 200.
3. if x equals your grade, what is the value of g to one decimal place?

>>...1. let f be a quadratic function...<<

f(x) = a(x-h)² + k

>>...with a minimum value of -22100,...<<

That tells us k = -22100, a > 0

So far we have

f(x) = a(x-h)² - 22100

>>...a line of symmetry of x=140, 

That tells us h = 140. So far we have

f(x) = a(x-140)² - 22100

>>...and a y-intercept of (0,-2500)...<<

That tells us f(0) = -2500. Substituting

-2500 = a(0-140)² - 22100 

19600 = 19600a

a = 1

So we now have

f(x) = 1(x-140)² - 22100

f(x) = (x-140)² - 22100

>>...2. the function g is the same as f but reflected about the
x-axis...<<

To reflect f(x) across the x-axis we multiply the right side
of its equation by -1

So far we have for the right side of g(x)

-[(x-140)² - 22100]

-(x-140)² + 22100 

>>...and reduced by a factor of 200...<<

That means to form the right side of g(x) we need to
multiply it by 1/200, or divide by 200:

[-(x-140)² + 22100]/200 or

-x²/200 + 7x/5 + 25/2

or

g(x) = -.005x² + 1.4x + 12.5

>>...3. if x equals your grade, what is the value of g to one
decimal place?...<<
You have to be given your grade.  Apparently this is a
teacher's function for curving grades.  Make a table of
values of possible grades, by substituting these values
of possible grades:  
If your grade is        Your curved grade is 
       x             g(x) = -.005x² + 1.4x + 12.5
      100            g(100) = 102.5
       90             g(90) = 98
       80             g(80) = 92.5
       70             g(70) = 86
       60             g(60) = 78.5
       50             g(50) = 70
       40             g(40) = 60.5
       30             g(30) = 50
       20             g(20) = 38.5
       10             g(10) = 26
        0              g(0) = 12.5        
   
If your grade is between two of these values, simply substitute
it for x in g(x) = -.005x² + 1.4x + 12.5 to get your curved grade.  

Edwin