Question: Transformation of Functions: Oct ’26 Name: Elsie Mzignwe
Big Idea 1: students have a conceptual understanding of mathematical situations
1. Given the graph of $y = f(x)$, identify any potential invariant point(s) under the transformation
$g(x) = -2f(x)$?
Invariant point $(-1,0)$
2. The graph of $y = f(x)$ is transformed into the graph $g(x) = -\frac{1}{2} f\left(\frac{1}{2} x + 2\right) - 1$. The
ordered pair $(4, 6)$ is found on the point of the original function $y = f(x)$. Explain
how you know the location of $(4, 6)$ after the function has been transformed and
identify its new location.
3. A function $f(x) = x^2 - x - 2$ is multiplied by a constant value $k$ to create a new function
$g(x) = kf(x)$. If the graph of $y = g(x)$ passes through point $(3, 14)$, state the value
of $k$.
Graph shape: piecewise linear graph in the top-left/center-left; it has open circles at approximately $(-4, 2)$, $(1, 1)$, and $(3, 1)$, with connected line segments passing through a filled vertex near $(-3, -2)$ and another filled point at $(-1, 0)$, then rising to a horizontal segment from about $x=1$ to $x=3$ at $y=1$. position_hint = top-left
4. Answer the following questions with respect to the graph that is found below:
y = $f(x)$
a) Without sketching, identify the domain and range of the inverse.
$D\{4 \leq y \leq -2, y \in \mathbb{R}\}$
$R\{ -6 \leq x \leq 2, x \in \mathbb{R}\}$
b) Will the inverse be a function? If not, restrict the domain on $f(x)$ so that the inverse is a function.
The y channel here then
said? Check here
to would then
Big Idea 2: students apply my mathematical knowledge to solve mathematical situations
1. Below is the function $y = \sqrt{x - 2}$.
$(2, 0)$
$(3, 1)$
$(6, 2)$
On the graph below, draw the inverse function. State the equation of the inverse functions.
Draw the inverse function and state the equation of this inverse. Include any restrictions that have been used so that the it looks like the function drawn below.
position_hint: top-left — graph labeled $y = f(x)$ is a piecewise line graph with open/closed endpoints at approximately $(-6,4)$, $(-2,4)$, $(0,-2)$, and $(2,3)$; graph labeled $y = \sqrt{x - 2}$ is a square-root curve starting at $(2,0)$ and passing through $(3,1)$ and $(6,2)$, extending rightward.
2. Given the function $f(x) = 3x + 5$, if $x = -25$ on the inverse function, explain how
you would determine the $y$ value on the inverse function. Why is this your method
of choice. NOTE: You do NOT need to solve this question mathematically.
3. Given the graph $y = f(x)$, sketch the graph of $y - 1 = 3f\left(- \frac{1}{2}(x - 4)\right)$
graph shape: line segment graph with connected points, position_hint=top-right
graph shape: piecewise linear graph with four marked points connected by line segments, position_hint=bottom-right
solve them in easy steps
1. **Problem 1: Identify invariant points under $g(x) = -2f(x)$ given $y = f(x)$ and invariant point $(-1,0)$.**
- An invariant point $(a,b)$ satisfies $g(a) = b$. Here, $g(x) = -2f(x)$, so $-2f(a) = b$.
- Since $f(a) = b$, substitute to get $-2b = b \Rightarrow -2b - b = 0 \Rightarrow -3b = 0 \Rightarrow b = 0$.
- The point $(-1,0)$ is invariant because $f(-1) = 0$ and $g(-1) = -2 \times 0 = 0$.
2. **Problem 2: Find new location of point $(4,6)$ under $g(x) = -\frac{1}{2} f\left(\frac{1}{2} x + 2\right) - 1$.**
- Given $f(4) = 6$. We want $g(x) = y$ such that $g(x) = -\frac{1}{2} f\left(\frac{1}{2} x + 2\right) - 1$.
- Set $\frac{1}{2} x + 2 = 4$ to find $x$ corresponding to $f(4)$.
- Solve: $\frac{1}{2} x + 2 = 4 \Rightarrow \frac{1}{2} x = 2 \Rightarrow x = 4$.
- Then $g(4) = -\frac{1}{2} f(4) - 1 = -\frac{1}{2} \times 6 - 1 = -3 - 1 = -4$.
- New location is $(4, -4)$.
3. **Problem 3: Find $k$ if $g(x) = kf(x)$ and $g(3) = 14$ with $f(x) = x^2 - x - 2$.**
- Calculate $f(3) = 3^2 - 3 - 2 = 9 - 3 - 2 = 4$.
- Since $g(3) = k f(3) = 14$, then $k \times 4 = 14 \Rightarrow k = \frac{14}{4} = \frac{7}{2} = 3.5$.
4. **Problem 4a: Domain and range of inverse of $y = f(x)$ given domain and range of $f$.**
- Domain of $f$: $-6 \leq x \leq 2$. Range of $f$: $4 \leq y \leq -2$ (likely a typo, assume $-2 \leq y \leq 4$).
- Domain of inverse $f^{-1}$ is range of $f$: $-2 \leq x \leq 4$.
- Range of inverse $f^{-1}$ is domain of $f$: $-6 \leq y \leq 2$.
5. **Problem 4b: Is inverse a function? Restrict domain if needed.**
- Inverse is a function if $f$ is one-to-one (passes horizontal line test).
- If not, restrict domain of $f$ to an interval where it is one-to-one, e.g., $[-6, -1]$ or $[-1, 2]$.
6. **Problem 5: Find inverse of $y = \sqrt{x - 2}$ and state equation and restrictions.**
- Start with $y = \sqrt{x - 2}$.
- Swap $x$ and $y$: $x = \sqrt{y - 2}$.
- Square both sides: $x^2 = y - 2$.
- Solve for $y$: $y = x^2 + 2$.
- Domain of original: $x \geq 2$, so range of inverse is $y \geq 0$.
- Restrict domain of inverse to $x \geq 0$ to match range of original.
- Inverse function: $f^{-1}(x) = x^2 + 2$, $x \geq 0$.
7. **Problem 6: Explain how to find $y$ on inverse of $f(x) = 3x + 5$ given $x = -25$ on inverse.**
- To find $y$ on inverse at $x = -25$, find $f(y) = -25$.
- Solve $3y + 5 = -25$ for $y$.
- This method uses the definition of inverse functions: $f^{-1}(x) = y$ means $f(y) = x$.
8. **Problem 7: Sketch graph of $y - 1 = 3f\left(- \frac{1}{2}(x - 4)\right)$.**
- This is a transformation of $f(x)$: horizontal shift right by 4, horizontal reflection and stretch by factor 2, vertical stretch by 3, vertical shift up by 1.
- Steps:
1. Replace $x$ by $-\frac{1}{2}(x - 4)$ inside $f$.
2. Multiply $f$ by 3.
3. Add 1 to entire function.
- Sketch accordingly based on original $f(x)$ graph.