SAT Math : Algebraic Functions
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Example Question #41 : Algebraic Functions
Define 
Determine 
First, we must determine whether 
A quadratic function has a parabola as its graph; this graph changes direction (downward to upward, or vice versa) at a given point called the vertex.
The 
is 
The 
The vertex of the graph of 
Example Question #1 : How To Find F(X)
If f(x) = x2 – 5 for all values x and f(a) = 4, what is one possible value of a?
14
1
3
8
11
3
Using the defined function, f(a) will produce the same result when substituted for x:
f(a) = a2 – 5
Setting this equal to 4, you can solve for a:
a2 – 5 = 4
a2 = 9
a = –3 or 3
Example Question #2 : How To Find F(X)
If the function g is defined by g(x) = 4x + 5, then 2g(x) – 3 =
4x + 2
8x + 2
6x + 7
8x + 7
6x + 2
8x + 7
The function g(x) is equal to 4x + 5, and the notation 2g(x) asks us to multiply the entire function by 2. 2(4x + 5) = 8x + 10. We then subtract 3, the second part of the new equation, to get 8x + 7.
Example Question #3 : Algebraic Functions
If f(x) = x2 + 5x and g(x) = 2, what is f(g(4))?
39
4
36
2
14
14
First you must find what g(4) is. The definition of g(x) tells you that the function is always equal to 2, regardless of what “x” is. Plugging 2 into f(x), we get 22 + 5(2) = 14.
Example Question #4 : How To Find F(X)
f(a) = 1/3(a3 + 5a – 15)
Find a = 3.
19
27
3
1
9
9
Substitute 3 for all a.
(1/3) * (33 + 5(3) – 15)
(1/3) * (27 + 15 – 15)
(1/3) * (27) = 9
Example Question #2 : Algebraic Functions
Evaluate f(g(6)) given that f(x) = x2 – 6 and g(x) = –(1/2)x – 5
58
–25
30
–8
50
58
Begin by solving g(6) first.
g(6) = –(1/2)(6) – 5
g(6) = –3 – 5
g(6) = –8
We substitute f(–8)
f(–8) = (–8)2 – 6
f(–8) = 64 – 6
f(–8) = 58
Example Question #1 : How To Find F(X)
If f(x) = |(x2 – 175)|, what is the value of f(–10) ?
–75
275
–275
15
75
75
If x = –10, then (x2 – 175) = 100 – 175 = –75. But the sign |x| means the absolute value of x. Absolute values are always positive.
|–75| = 75
Example Question #3 : How To Find F(X)
If f(x)= 2x² + 5x – 3, then what is f(–2)?
–21
7
–5
–1
–5
By plugging in –2 for x and evaluating, the answer becomes 8 – 10 – 3 = -5.
Example Question #3 : Algebraic Functions
If f(x) = x² – 2 and g(x) = 3x + 5, what is f(g(x))?
3x² – 1
9x² + 30x + 23
9x² + 30x + 25
9x² + 23
9x² + 30x + 23
To find f(g(x) plug the equation for g(x) into equation f(x) in place of “x” so that you have: f(g(x)) = (3x + 5)² – 2.
Simplify: (3x + 5)(3x + 5) – 2
Use FOIL: 9x² + 30x + 25 – 2 = 9x² + 30x + 23
Example Question #7 : How To Find F(X)
f(x) = 2x2 + x – 3 and g(y) = 2y – 7. What is f(g(4))?
42
0
-33
57
33
0
To evaluate f(g(4)), one must first determine the value of g(4), then plug that into f(x).
g(4) = 2 x 4 – 7 = 1.
f(1) = 2 x 12 + 2 x 1 – 3 = 0.
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