Question 526·Hard·Math

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The graph of $y=f(x)$ is a parabola. Its x-intercepts have x-coordinates $-1$ and $k$ , where $k>-1$ . The axis of symmetry is $x=2$ , and the y-coordinate of the vertex is $-18$ . The function $g$ is defined by $g(x)=f(3-x)+4$ . Which choice gives an equation for $g$ ?

A quadratic transformation question gives features of one parabola, then defines a new function from it. Use the intercepts and their midpoint to build the original rule, and use the vertex height to find its missing coefficient. Then define both functions in Desmos and compare the new graph with the choices. The key trap is reading $3-x$ as $x-3$ .

Hints

Hint 1
The axis of symmetry is the mirror line halfway between a parabola’s two $x$ -intercepts. The intercept at $-1$ is $3$ units left of $x=2$ . Where must the other intercept be?
Hint 2
An $x$ -intercept makes the function equal $0$ , so each intercept gives a factor. The vertex supplies an output at $x=2$ ; use that output to find the coefficient in front of the factors.
Hint 3
In $g(x)=f(3-x)+4$ , the entire $3-x$ is the input to $f$ . Define $f$ first in Desmos, then enter the given rule for $g$ to see the new parabola.

Step-by-step

Approach 1: Build $f$ , then graph $g$
Step 1
Find the second intercept
The axis of symmetry mirrors the two intercepts. Since $-1$ is $3$ units left of $2$ , the other intercept is $3$ units right:
$k=2+3=5.$
Mirror across $x=2$ , not across the $y$ -axis.
Step 2
Write a rule with those intercepts
An $x$ -intercept is an input where $f(x)=0$ . The factor $x+1$ is $0$ at $-1$ , and $x-5$ is $0$ at $5$ . So use factored form:
$f(x)=a(x+1)(x-5).$
The coefficient $a$ still needs to be found; the intercepts alone don't fix how steep the parabola is.
Step 3
Turn the vertex height into an equation
The vertex is where the parabola turns. It lies on the axis $x=2$ , and its given output is $-18$ , so $f(2)=-18$ . Substitute $2$ into the rule:
$-18=a(2+1)(2-5).$
Multiply the two parentheses:
$-18=-9a.$
An intercept wouldn't find $a$ : there, both sides would be $0$ for any value of $a$ .
Step 4
Find the coefficient in Desmos
Type $-18\sim a(2+1)(2-5)$ . The $\sim$ tells Desmos to find the value of $a$ that fits the vertex equation. Under PARAMETERS, Desmos shows $a=2$ .
Step 5
Draw the transformed parabola
Type $f(x)=a(x+1)(x-5)$ , then type the given rule $g(x)=f(3-x)+4$ . Click the vertex of the new curve: Desmos shows $(1,-14)$ . At the new input $1$ , the whole inside is $3-1=2$ , reaching $f$ 's vertex; the outside $+4$ raises its output from $-18$ to $-14$ . Match the whole input inside $f$ first, then change the output outside.
Step 6
Match the whole graph
Graph the candidate $y=2(x+2)(x-4)+4$ on the next line. It overlaps $g$ across the curve, not at only one point, so it gives the same output for every input. The equation for $g$ is $g(x)=2(x+2)(x-4)+4$ . Choice B.

Approach 2: See the match by substitution
Step 1
Replace each input in $f$
Using the rule found for $f$ , replace $x$ in both factors with the whole input $3-x$ :
$g(x)=2((3-x)+1)((3-x)-5)+4.$
Simplify inside the factors:
$g(x)=2(4-x)(-2-x)+4.$
Step 2
Cancel the two negative signs
Each factor has a negative sign. Rewrite both before multiplying, so the two negatives cancel:
$g(x)=2[-(x-4)][-(x+2)]+4.$
Multiply the two negative signs:
$g(x)=2(x-4)(x+2)+4.$
That is the matching equation for $g$ . Choice B.

Question 526·Hard·Math

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