Question

apply your knowledge of reflections to answer each question below.
a
use the graph of triangle jkl to answer 1-2.

1. reflect jkl over the x-axis. record the coordinates of the image below.

j
k -5,-2
l -2,-3

1. write an algebraic representation for the reflection.

(x,y)→x,—y)

b
the table represents the location of qrst before and after a reflection. use the table to answer 3-5.
pre-image
image
q(-9, -8)
q(9, -8)
r(-9, -2)
r(9, -2)
s(-4, -2)
s(4, -2)
t(-4, -8)
4 ?-8

1. give a verbal description of the transformation.
2. which best represents the transformation algebraically?

a. (x, -y)
b. (-x, y)
c. (-x, -y)
d. (x + 18, y

1. find the location of t. 4,-8

Explanation:

Step1: Analyze Reflection Over x - axis

To reflect a point \((x,y)\) over the \(x\) - axis, the rule is that the \(x\) - coordinate remains the same and the \(y\) - coordinate changes its sign. So the transformation is \((x,y)\to(x, - y)\).

Step2: Verify with Given Points (for part A)

For example, if we have a point \(K\) with coordinates (let's assume from the graph, though we can use the general rule). If the original \(K\) has coordinates \((- 5,2)\) (from the graph's upper part), after reflection over \(x\) - axis, \(K'\) will be \((-5,-2)\), which matches the given answer for \(K'\). Similarly, for \(L\) with original coordinates \((-2,3)\), \(L'\) is \((-2,-3)\). And for \(J\) (assuming original \(J\) is \((-7,7)\)), \(J'\) is \((-7,-7)\).

Step3: Algebraic Representation (part 2 of A)

The algebraic representation for reflection over the \(x\) - axis is \((x,y)\to(x, - y)\) because when we reflect over the \(x\) - axis, the horizontal (x) position doesn't change, but the vertical (y) position is mirrored across the \(x\) - axis (so positive \(y\) becomes negative and vice - versa).

Step4: Analyze Reflection for part B (Q, R, S, T)

Looking at the pre - image and image coordinates: For \(Q(-9,-8)\) and \(Q'(9,-8)\), the \(x\) - coordinate changes from \(-9\) to \(9\) (i.e., \(-x\) where \(x = - 9\) gives \(9\)) and the \(y\) - coordinate remains \(-8\). Similarly for \(R(-9,-2)\to R'(9,-2)\) and \(S(-4,-2)\to S'(4,-2)\). So this is a reflection over the \(y\) - axis, and the algebraic rule is \((x,y)\to(-x,y)\). For \(T(-4,-8)\), using the rule \((x,y)\to(-x,y)\), \(T'\) will be \((4,-8)\).

Answer:

1. \(J'(-7,-7)\), \(K'(-5,-2)\), \(L'(-2,-3)\)
2. \((x,y)\to(x, - y)\)
3. The transformation is a reflection over the \(y\) - axis (since the \(x\) - coordinates of the pre - image and image are negatives of each other and \(y\) - coordinates are the same).
4. b. \((-x,y)\)
5. \(T'(4,-8)\)