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