Question 606·Medium·Math

0 / 2000

The quadratic function $q$ is defined by

q(x) = (x - 4)(x + 2).

A new function is defined by $r(x) = q(x - 3)$ . At which of the following $x$ -coordinates does the graph of $y = r(x)$ intersect the $x$ -axis at its greater $x$ -intercept?

A rule like $r(x)=q(x-h)$ signals a horizontal shift: the new graph keeps the same outputs but reaches them at different $x$ -values. For an intercept question, graph the original and shifted function in Desmos, then click where the shifted graph meets the $x$ -axis. Subtracting inside moves the graph right, not left. If there are two crossings, use the one the question requests.

Hints

Hint 1
An $x$ -intercept is a point where the graph meets the horizontal axis, so its output is $0$ . A quadratic may have two. What crossings does Desmos show for the original function?
Hint 2
In $q(x-3)$ , the whole input to $q$ is $x-3$ . To get an old input $u$ , the new coordinate must satisfy $x-3=u$ . Does that move each crossing left or right?

Step-by-step

Approach 1: Graph the shifted function
Step 1
Find the original crossings
Type $q(x)=(x-4)(x+2)$ in Desmos and click its marked $x$ -intercepts, where the output is $0$ . Desmos shows $(-2,0)$ and $(4,0)$ . These points belong to the original graph.
Step 2
Work out which way the graph moves
If $u$ is an old input to $q$ , the new graph gets that same output where $x-3=u$ . Add $3$ to both sides: $x=u+3$ . So subtracting $3$ inside moves each crossing $3$ units right, not left.
Step 3
Read the greater new intercept
Add $r(x)=q(x-3)$ beneath $q$ in Desmos and click the shifted curve's marked $x$ -intercepts. Desmos shows $(1,0)$ and $(7,0)$ , each $3$ units right of an original crossing. The greater $x$ -coordinate is $7$ , so $y=r(x)$ meets the $x$ -axis there. Choice D.

Approach 2: Find the zeros from the factors
Step 1
Put the new input into both factors
In $r(x)=q(x-3)$ , the entire $x-3$ takes the place of $x$ in both factors of $q$ :
$r(x)=((x-3)-4)((x-3)+2).$
Step 2
Simplify the factors
Combine the constants in each factor:
$r(x)=(x-7)(x-1).$
Step 3
Set the output to zero
At an $x$ -intercept, $r(x)=0$ . The zero-product property says a product is $0$ when at least one factor is zero, so:
$x-7=0\quad\text{or}\quad x-1=0.$
Step 4
Choose the greater root
Add $7$ to the first equation: $x=7$ . Add $1$ to the second equation: $x=1$ . The greater root is the greater $x$ -intercept of $r$ , at $x=7$ . Choice D.

Question 606·Medium·Math

Strategy

Hints

Step-by-step

Lessons that teach this

Study Hints

Strategy

Hints

Step-by-step

Lessons that teach this