Question 811·Hard·Math

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The rational function $f$ is defined by

f(x)=\frac{a}{x+b}+c,

where $a$ , $b$ , and $c$ are constants. The partial graph of $y=f(x)$ is shown.

A new function is defined by

g(x)=2f(2x-2)-3.

Which choice could define function $g$ ?

A function transformation changes a graph’s inputs and outputs separately. A labeled point gives one exact input-output pair, so you can test the choices without finding the entire rule for $f$ . Set the expression inside $f$ equal to the old input before changing its output. Then evaluate each choice at that same new input in Desmos. Guessing a shift from $2x-2$ can send you to the wrong input.

Hints

Hint 1
A graph point $(u,v)$ means $f(u)=v$ : input $u$ gives output $v$ . Read the labeled point on $f$ first. What exact input-output pair does it give you?
Hint 2
In a function transformation, the expression inside $f$ must equal the old input to recover its output. Set $2x-2$ equal to the input from the labeled point. What new $x$ works?
Hint 3
The output changes after you find the new input: multiply the old output by $2$ , then subtract $3$ . Evaluate the proposed rules at the same new input to see which one agrees.

Step-by-step

Approach 1: Move one point and test the choices

Step 1
Read the known output
The labeled point is $(1,0)$ . A graph point $(x,y)$ on $y=f(x)$ means input $x$ gives output $y$ , so $f(1)=0$ . You only need this one point to test the proposed rules.
Step 2
Find the new input for that point
To use $f(1)$ in $g$ , the whole expression inside $f$ must equal $1$ . Type $2x_1-2\sim1$ in Desmos. The $\sim$ asks it to find the value of $x_1$ that makes the sides equal. Under PARAMETERS, Desmos shows $x_1=1.5$ , so the old input $1$ lands at the new input $1.5$ .
Step 3
Find the new output
At $x=1.5$ , the input to $f$ is $1$ , whose output is $0$ . Type $2(0)-3$ ; Desmos shows $-3$ . So the transformed point is $(1.5,-3)$ . Don’t use $f(1.5)=0$ : the known output $0$ belongs to the old input $1$ .
Step 4
Test each proposed rule
Keep Desmos’s $x_1=1.5$ and type the four proposed expressions in order, replacing their $x$ with $x_1$ . Desmos shows $-3$ , $-1$ , $-15$ , and $5$ . Only the first expression gives $g(1.5)=-3$ , so it could define $g$ : $g(x)=-\frac{8}{2x-1}+1$ . Choice A.

Approach 2: Build the exact rule

Step 1
Read the constants from the dashed lines
An asymptote is a line the curve approaches. The horizontal asymptote $y=2$ gives $c=2$ . The vertical asymptote occurs where $x+b=0$ , or $x=-b$ . Since that dashed line is at $x=-1$ , $b=1$ , not $-1$ .
Step 2
Use the labeled point to set up an equation
The point $(1,0)$ says $f(1)=0$ . Put input $1$ and output $0$ into $f(x)=\frac{a}{x+1}+2$ to find the remaining constant: $0=\frac{a}{2}+2$ .
Step 3
Find the remaining constant
Type $a/2+2\sim0$ in Desmos. Under PARAMETERS, it shows $a=-4$ . So the rule read from the graph is $f(x)=-\frac{4}{x+1}+2$ .
Step 4
Substitute the whole new input
In $g(x)=2f(2x-2)-3$ , every $x$ in $f$ becomes $2x-2$ . Keep the outside changes in place: $g(x)=2\left(-\frac{4}{(2x-2)+1}+2\right)-3$ .
Step 5
Combine the denominator terms
Add $-2$ and $1$ inside the denominator: $g(x)=2\left(-\frac{4}{2x-1}+2\right)-3$ . Keeping the $2x$ matters; dropping its coefficient changes the rule.
Step 6
Multiply the outside factor
Distribute the outside $2$ to both terms in parentheses: $g(x)=-\frac{8}{2x-1}+4-3$ .
Step 7
Finish the rule
Subtract $3$ from the constant term: $g(x)=-\frac{8}{2x-1}+1$ . This is the exact transformed function. Choice A.

Question 811·Hard·Math

Strategy

Hints

Step-by-step

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Strategy

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Step-by-step

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