Example Problems

Example 1

In the function (y=3x+2), when (x=2), what is (y)?

Pretty easy, right?  We substitute 2 for x, and solve to get 8.  

Example 2

In the function (f(q)), where (f(q)) is equal to (3q+2), what is (f(2))?

It’s the exact same thing, made to look more confusing.  Our (f(q)) is the same as y, and “is equal to” is the same as the = operator.  Whatever is in the parenthesis we plug in for our variable.

$$y=3left(2right)+2=8$$

Example 3

Sometimes you might get the f ( ) and variable parts, and need to figure out the equation.

If (3,2) is a solution, which of the following could represent the function?

A) (y=3x+2)
B) (y=2x+3)
C) (y=x)
D) (y=x-1)

Here we are given a coordinate pair, which represents our x and y.  All we have to do is see which equation gives us a correct equality when we plug them in.  D) is the only one that returns 2 when 3 is plugged in for x.

Example 4

$$g(x)=ax^2+24$$

For the function defined above, is a constant and g(4) = 8. What is the value of g(−4) ? 

Here we have two things that are missing:  a constant and a variable.  We can’t solve a single equation with two things missing, so we first need to find a with the information they gave us before – an x,y pair in the form of g(4) = 8.

$$8=a(4)^2+24\8=a16+24\-16=a16\a=-1$$

Now that we have our a, so we can plug in that and our -4.

$$y=-1(-4)^2+24\y=-16+24=8$$

Example 5

If (fleft(xright)=x^2-x+3) and (gleft(xright)=-x+2) what is (f(gleft(xright))) for (g(3))?

For nested functions, we always start in the innermost parentheses.  We take what we’re given for g inside the parentheses, and plug into the equation.

$$-3+2=-1$$

So by plugging in 3, g(x) returns -1.  Now we take that returned value, and plug it into (f(x)).

$${-1}^2-left(-1right)+3=1+1+3=5$$

Example 6

A function f satisfies (fleft(2right)=3) and (fleft(3right)=5).  A function g satisfies (gleft(3right)=2) and (gleft(5right)=6).  What is (f(gleft(3right)))?

A) 2
B) 3
C) 5
D) 6

This is a variant of example 5.  We have nested functions, but the functions aren’t given.  We only have two points, so we can’t figure out what the functions are (could be linear or quadratic), but fortunately, we don’t have to.  Notice that we need to start with (g(3)) since it is innermost.  They tell us what (g(3)) returns!  When you plug in 3, that function returns 2.  Now we need to take 2 into our (f) function – they give us that too!  (f(2)=3).  We’re all done.  The answer is B).

Example 7

$$f(x)={(x-b)}^2$$

The function f is defined by the equation above, where b is a constant. If (f(5)=f(9)), what is the value of (f(1))?

A) 25
B) 36
C) 45
D) 49

Here, we are told that the function is equal when we plug in two different numbers.  So, all we have to do is plug in both numbers, set them equal, and solve for b.  Once we have b, we can plug in 1 and then find our (x), or y, value.

$$(x-b)^2=x^2-2bx+b^2\5^2-2b5+b^2=9^2-2b9+b^2\25-10b+b^2=81-18b+b^2\8b=56\b=7$$

So then:

$$(1-7)^2=36$$

Example 8

Here’s a particularly awful problem from a previous test.  Remember the rules discussed above.

$$f(x)=(x+6)(x-4)$$

Which of the following is an equivalent form of the function above in which the minimum value of appears as a constant or coefficient? 

Ugh, what?  Let’s read it again.  We need the minimum value of f That is, we need the smallest value of y that this function can return.  Then, we need to see that value appear in our equation as a coefficient to an or (x^2) term, or as just a regular number hanging out by itself (constant).

Notice this is a quadratic, and because the (x^2) coefficient (here, 1) is positive, it will open upwards.  That means the minimum value of y will be found at the vertex. 

How do we figure out what that is?  Remember that the vertex is always halfway in between the two points where the parabola crosses the x-axis.  Here, we have solutions at -6 and 4, so our midpoint would be where x = -1.  Plugging this in reveals that y would be -25, so we just have to go find that in the answers as a constant or coefficient.

Alternatively, you might start by FOILing the original equation.  This gives you answer B), so we’ll keep that as a possibility for now.  We see answer A) is wrong because it is NOT an equivalent expression, it is missing the term.  We can cross this out.  The smallest value in B) expressed as a coefficient is 1.  The smallest constant is -24.  Looking at C), there is nothing smaller than -24 so we cross that one off without doing anything else with it.

D) has a smaller number as a constant, -25.  This looks promising!  But how do we confirm?  This is the hardest part of this problem.  You have to recognize that D) is written in vertex form of a parabola,

$$y=a(x-h)^2+k$$

If you FOIL that out D), you get our original equation, proving that they are equivalent and it has the smallest constant.