This image contains a set of problems dealing with transformations, reflections, and dilations of functions.

1. Translations

You are given functions and translation vectors. To apply a translation to a function, the following rules apply:

• Horizontal translation $f(x + c)$ shifts the function by $c$ units in the opposite direction of the sign.
• Vertical translation $f(x) + d$ shifts the function by $d$ units up or down.

a) $f(x) = 3x^2 - x - 2$ and translation $[-1, -2]$:

• The horizontal translation by $-1$: Replace $x$ with $x + 1$.
• The vertical translation by $-2$: Subtract 2 from the function. $g(x) = f(x + 1) - 2$ First, compute $f(x + 1)$: $f(x + 1) = 3(x + 1)^2 - (x + 1) - 2 = 3(x^2 + 2x + 1) - x - 1 - 2 = 3x^2 + 6x + 3 - x - 3 - 2 = 3x^2 + 5x - 2$ Now apply the vertical translation: $g(x) = 3x^2 + 5x - 2 - 2 = 3x^2 + 5x - 4$

b) $f(x) = 3x^2 - x - 2$ and translation $[-4, -1]$:

• Horizontal translation by $-4$: Replace $x$ with $x + 4$.
• Vertical translation by $-1$: Subtract 1 from the function. $g(x) = f(x + 4) - 1$ First, compute $f(x + 4)$: $f(x + 4) = 3(x + 4)^2 - (x + 4) - 2 = 3(x^2 + 8x + 16) - x - 4 - 2 = 3x^2 + 24x + 48 - x - 6 = 3x^2 + 23x + 42$ Now apply the vertical translation: $g(x) = 3x^2 + 23x + 42 - 1 = 3x^2 + 23x + 41$

2. Reflections

Reflections can occur over the x-axis, y-axis, or other lines.

a) $f(x) = 3x^2 + 1$ and reflection over $x = -2$:

For a reflection over $x = -2$, the transformation involves shifting $x$ and reflecting. Replace $x$ with $-x - 4$. $g(x) = f(-x - 4) = 3(-x - 4)^2 + 1 = 3(x^2 + 8x + 16) + 1 = 3x^2 + 24x + 48 + 1 = 3x^2 + 24x + 49$

b) $f(x) = 4x + 6$ and reflection over $y = -x$:

To reflect over the line $y = -x$, swap $x$ with $-y$ in the equation. $g(x) = -f(-x) = -(4(-x) + 6) = -(-4x + 6) = 4x - 6$

3. Dilations

Dilation involves scaling the function vertically or horizontally.

a) $f(x) = 6x^2 - 4x + 2$ and vertical scaling by $\frac{1}{2}$:

To vertically scale the function, multiply the entire function by $\frac{1}{2}$: $g(x) = \frac{1}{2}f(x) = \frac{1}{2}(6x^2 - 4x + 2) = 3x^2 - 2x + 1$

b) $f(x) = 2x^2 + 3x - 4$ and horizontal scaling by 2:

For horizontal scaling, replace $x$ with $\frac{x}{2}$ in the function: $g(x) = f\left(\frac{x}{2}\right) = 2\left(\frac{x}{2}\right)^2 + 3\left(\frac{x}{2}\right) - 4 = 2\left(\frac{x^2}{4}\right) + 3\left(\frac{x}{2}\right) - 4 = \frac{x^2}{2} + \frac{3x}{2} - 4$

c) $f(x) = ?$ and vertical scaling by 2 for $g(x) = 4x - 2$:

To get $f(x)$ from $g(x) = 4x - 2$ and vertical scaling by 2, divide the function by 2: $f(x) = \frac{1}{2}(4x - 2) = 2x - 1$

Let me know if you need further details or explanation.

5 Related Questions:

1. How do you apply horizontal scaling in function transformations?
2. What changes occur to a function when it is reflected over the y-axis?
3. Can you explain how to reflect a function over a vertical line other than the y-axis?
4. How does vertical dilation affect the graph of a quadratic function?
5. What is the difference between horizontal and vertical translations in terms of function graphs?

Tip: When transforming functions, remember that horizontal transformations (translations and dilations) affect the input $x$, while vertical transformations affect the output of the function.