Transformation of Functions: Translation

Vertical Translation

For any function $f(x)$ and $c>0$, $f(x)+c$ vertically shifts the graph of $f(x)$ upward by $c$ units and $f(x)-c$ vertically shifts the graph of $f(x)$ downward by $c$ units.

Horizontal Translation

For any function $f(x)$ and $c>0$, $f(x-c)$ horizontally shifts the graph of $f(x)$ right by $c$ units and $f(x+c)$ horizontally shifts the graph of $f(x)$ left by $c$ units.

Question (1)

Use the graph of the function $f$ to sketch the graph of the following functions.


$\begin{array}{lll} \text{(a) } g(x) = f(x) + 1 \\\\ \text{(b) } h(x) = f(x)-1 \\\\ \text{(c) } p(x) = f(x-1) \\\\ \text{(d) } F(x) = f(x+2) \\\\ \text{(e) } G(x) = f(x+1) - 2\\\\ \text{(f) } H(x) = f(x-1) + 1 \end{array}$

(a) $g(x) = f(x) + 1$ The grph of $y=g(x)$ can be obtained by shifting the grph of $y=f(x)$ 1 unit up. (b) $h(x) = f(x)-1$ The grph of $y=h(x)$ can be obtained by shifting the grph of $y=f(x)$ 1 unit down. (c) $p(x) = f(x-1)$ The grph of $y=p(x)$ can be obtained by shifting the grph of $y=f(x)$ 1 unit right. (d) $F(x) = f(x+2)$ The grph of $y=F(x)$ can be obtained by shifting the grph of $y=f(x)$ 2 units left. (e) $G(x) = f(x+1)$ -2 The grph of $y=G(x)$ can be obtained by shifting the grph of $y=f(x)$ 1 unit left followed by 2 units down. (f) $H(x) = f(x-1)$ + 1 The grph of $y=H(x)$ can be obtained by shifting the grph of $y=f(x)$ 1 unit right followed by 1 unit up.

Question (2)

Draw the graph $f(x) = x^2$. Hence using the transformations of $f(x)$, draw the graph of the following functions.

$\begin{array}{lll} \text{(a) } g(x) = f(x) + 1 \\\\ \text{(b) } h(x) = f(x) - 3 \\\\ \text{(c) } p(x) = f(x-1) \\\\ \text{(d) } F(x) = f(x+3) \\\\ \text{(e) } G(x) = f(x+1) - 2 \\\\ \text{(f ) } H(x) = f(x-2) + 3 \end{array}$

To sketch the graph of $y=x^2$, we should find some sample points on the graph. $\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|} \hline x & -2 & -1 & 0 & 1 & 2\\ \hline y & 4 & 1 & 0 & 1 & 4\\ \hline \end{array}$ (a) $g(x)=f(x)+1$ The grph of $y=g(x)$ can be obtained by shifting the grph of $y=f(x)$ 1 unit up. (b) $h(x)=f(x)−3$ The grph of $y=h(x)$ can be obtained by shifting the grph of $y=f(x)$ 3 units down. (c) $p(x)=f(x−1)$ The grph of $y=p(x)$ can be obtained by shifting the grph of $y=f(x)$ 1 unit right. (d) $F(x)=f(x+3)$ The grph of $y=F(x)$ can be obtained by shifting the grph of $y=f(x)$ 3 units left. (e) $G(x)=f(x+1)−2$ The grph of $y=G(x)$ can be obtained by shifting the grph of $y=f(x)$ 1 unit left followed by 2 units down. (f) $H(x)=f(x−2)+3$ The grph of $y=H(x)$ can be obtained by shifting the grph of $y=f(x)$ 2 units right followed by 3 units up.

Question (3)

Draw the graph $f(x) = |x|$. Hence using the transformations of $f(x)$, draw the graph of the following functions.

$\begin{array}{lll} \text{(a) } g(x) = f(x) + 1 \\\\ \text{(b) } h(x) = f(x) - 3 \\\\ \text{(c) } p(x) = f(x-1) \\\\ \text{(d) } F(x) = f(x+3) \\\\ \text{(e) } G(x) = f(x+1) - 2 \\\\ \text{(f ) } H(x) = f(x-2) + 3 \end{array}$

To sketch the graph of $y=x^2$, we should find some sample points on the graph. $\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|} \hline x & -2 & -1 & 0 & 1 & 2\\ \hline y & 2 & 1 & 0 & 1 & 2\\ \hline \end{array}$ (a) $g(x)=f(x)+1$ The grph of $y=g(x)$ can be obtained by shifting the grph of $y=f(x)$ 1 unit up. (b) $h(x)=f(x)−3$ The grph of $y=h(x)$ can be obtained by shifting the grph of $y=f(x)$ 3 units down. (c) $p(x)=f(x−1)$ The grph of $y=p(x)$ can be obtained by shifting the grph of $y=f(x)$ 1 unit right. (d) $F(x)=f(x+3)$ The grph of $y=F(x)$ can be obtained by shifting the grph of $y=f(x)$ 3 units left. (e) $G(x)=f(x+1)−2$ The grph of $y=G(x)$ can be obtained by shifting the grph of $y=f(x)$ 1 unit left followed by 2 units down. (f) $H(x)=f(x−2)+3$ The grph of $y=H(x)$ can be obtained by shifting the grph of $y=f(x)$ 2 units right followed by 3 units up.

Question (4)

Draw the graph $f(x) = \sqrt{x}$. Hence using the transformations of $f(x)$, draw the graph of the following functions.

$\begin{array}{lll} \text{(a) } g(x) = f(x) + 1 \\\\ \text{(b) } h(x) = f(x) - 3 \\\\ \text{(c) } p(x) = f(x-1) \\\\ \text{(d) } F(x) = f(x+3) \\\\ \text{(e) } G(x) = f(x+1) - 2 \\\\ \text{(f ) } H(x) = f(x-2) + 3 \end{array}$

To sketch the graph of $y=x^2$, we should find some sample points on the graph. $\begin{array}{|c|c|c|c|} \hline x & 0 & 1 & 4 \\ \hline y & 0 & 1 & 2 \\ \hline \end{array}$ (a) $g(x)=f(x)+1$ The grph of $y=g(x)$ can be obtained by shifting the grph of $y=f(x)$ 1 unit up. (b) $h(x)=f(x)−3$ The grph of $y=h(x)$ can be obtained by shifting the grph of $y=f(x)$ 3 units down. (c) $p(x)=f(x−1)$ The grph of $y=p(x)$ can be obtained by shifting the grph of $y=f(x)$ 1 unit right. (d) $F(x)=f(x+3)$ The grph of $y=F(x)$ can be obtained by shifting the grph of $y=f(x)$ 3 units left. (e) $G(x)=f(x+1)−2$ The grph of $y=G(x)$ can be obtained by shifting the grph of $y=f(x)$ 1 unit left followed by 2 units down. (f) $H(x)=f(x−2)+3$ The grph of $y=H(x)$ can be obtained by shifting the grph of $y=f(x)$ 2 units right followed by 3 units up.

Question (5)

If the graph of the quadratic function $f(x)$ has the vertex at $(-1,3)$, state the vertex after the given translations:

$\begin{array}{lll} \text{(a) } f(x-2)+2\\\\ \text{(b) } f(x)-5 \\\\ \text{(c) } f(x+1)-3 \\\\ \text{(d) } f(x-6) \\\\ \text{(e) } f(x+1) - 2\qquad\quad \\\\ \text{(f) } f(x+2) + 1 \end{array}$

$\begin{aligned} \text{(a) } & f(x-2)+2 \\\\ & (-1, 3)\rightarrow (-1+2, 3+2)=(1, 5)\\\\ \text{(b) } & f(x)-5 \\\\ & (-1, 3)\rightarrow (-1, 3-5)=(-1, -2)\\\\ \text{(c) } & f(x+1)-3 \\\\ & (-1, 3)\rightarrow (-1-1, 3-3)=(-2, 0)\\\\ \text{(d) } & f(x-6) \\\\ & (-1, 3)\rightarrow (-1+6, 3)=(5, 3)\\\\ \text{(e) } & f(x+1)-2 \\\\ & (-1, 3)\rightarrow (-1-1, 3-2)=(-2, 1)\\\\ \text{(f) } & f(x+2)+1 \\\\ & (-1, 3)\rightarrow (-1-2, 3+1)=(-3, 4) \end{aligned}$

Question (6)

If the points $A(2,-3)$ lies on the graph of $y=f(x)$. Use transformation to find the map point of $A$ on the graph $y= g(x)$ such that

(a) $g(x) = f(x-2) + 1$.

(b) $g(x) = f(x+1) - \frac{1}{2}$.

Let the mapped point of $A(2, -3)$ be $A'(a,b)$. $\begin{aligned} &\\ \text{(a) } & \text{ After translation by } f(x-2)+1 \\\\ & a=2+2 = 4, b=-3+1=2\\\\ & \text{ The mapped point is } A'(4,2)\\\\ \text{(b) } & \text{ After translation by } f(x+1) -\frac{1}{2} \\\\ & a=2-1 = 1, b=-3-\frac{1}{2}=- \frac{7}{2}\\\\ & \text{ The mapped point is } A'\left(1,-\frac{7}{2}\right)\\\\ \end{aligned}$

Question (7)

State the parent function and the translation that is occurring in each of the following functions.

စာဖတ်သူ၏ အမြင်ကို လေးစားစွာစောင့်မျှော်လျက်!