Free Math Worksheets

- How do the coordinates of a point change under reflection across the y-axis or x-axis? - Translate those coordinate changes into changes in the function rule. - What must be true about the slopes of two parallel lines?

1. Reflecting across the y-axis replaces \(x\) with \(-x\): \(h_1(x) = -1.5(-x) + 6 = 1.5x + 6\). 2. Reflecting across the x-axis multiplies each output by \(-1\): \(h_2(x) = -(-1.5x + 6) = 1.5x - 6\). 3. Both image lines have slope \(1.5\). Since their y-intercepts are different, they are distinct parallel lines.

a) \(h_1(x) = 1.5x + 6\) b) \(h_2(x) = 1.5x - 6\) c) Yes. Both lines have slope \(1.5\).

- Find the slope from the two given points. - Which points stay fixed under a reflection across \(y = 2\)? - Reflect the origin by using its distance from \(y = 2\). - Compare the horizontal and vertical changes before and after the reflection.

1. The slope through \((0, 0)\) and \((4, 2)\) is \(\frac{2}{4} = 0.5\), so \(k(x) = 0.5x\). 2. Point \(B(4, 2)\) lies on the line of reflection, so it stays fixed. The origin is 2 units below \(y = 2\), so it reflects to \((0, 4)\). 3. The image line through \((0, 4)\) and \((4, 2)\) has slope \(\frac{2 - 4}{4 - 0} = -0.5\), so \(k'(x) = -0.5x + 4\). 4. A horizontal reflection keeps horizontal changes the same but reverses vertical changes. Therefore, \(\frac{\Delta y}{\Delta x}\) changes sign.

a) \(k(x) = 0.5x\) b) \(k'(x) = -0.5x + 4\) c) The reflection reverses \(\Delta y\) while keeping \(\Delta x\) unchanged, so the slope changes sign.

- Compare the input \(6-x\) with the general reflected input \(2a-x\). - A reflection line bisects the segment joining a point and its image. - Use the midpoint formula on the number line. - Determine the constant midpoint required for reflection across \(x=a\).

1. Since \(h(x)=f(6-x)\), compare \(6-x\) with \(2a-x\). This gives \(2a=6\), so \(a=3\). 2. For any input \(x\), let its reflected input be \(x'=2a-x\). 3. Their midpoint is \(\frac{x+x'}{2}=\frac{x+(2a-x)}{2}=a\). 4. Also, \(|x-a|=|x'-a|\). Thus, \(x\) and \(x'\) lie the same distance from \(x=a\) on opposite sides, so the transformation is a reflection across that line.

The reflection line is \(x=3\). For \(x'=2a-x\), the midpoint is \(\frac{x+x'}{2}=a\), and the two inputs are equidistant from \(x=a\).

- Reflection preserves distance to points on the mirror line. - Replace the broken path to \(B\) with a path to its reflected point. - The shortest distance between two points is a straight segment. - Use the triangle inequality to compare any other point on the bank.

1. Reflect \(B\) across line \(g\) to point \(B'\). 2. For every point \(X\) on \(g\), reflection preserves distance and fixes \(X\), so \(XB = XB'\). 3. Therefore, \(AX + XB = AX + XB'\). 4. Draw \(\overline{AB'}\). Let its intersection with \(g\) be \(T\). Since \(A\), \(T\), and \(B'\) are collinear, \(AT + TB = AT + TB' = AB'\). 5. For any other point \(Q\) on \(g\), the triangle inequality gives \(AQ + QB' > AB'\). 6. Since \(QB' = QB\), \(AQ + QB > AT + TB\). Thus, \(T\) is the unique minimizing location.

Reflect \(B\) across \(g\) to \(B'\). Place the water station at \(T = \overline{AB'} \cap g\). This location minimizes \(AT + TB\).