Question

1. the graph of efgh is shown. graph the image of efgh after a translation of 4 units left and 6 units up. write the coordinates of the image.

Explanation:

Step1: Recall translation rule

For a point $(x,y)$ translated 4 units left and 6 units up, the new - point $(x',y')$ has coordinates $x'=x - 4$ and $y'=y + 6$.

Step2: Assume coordinates of vertices

Suppose the coordinates of the vertices of $EFGH$ are $E(x_E,y_E)$, $F(x_F,y_F)$, $G(x_G,y_G)$, $H(x_H,y_H)$. From the graph (assuming $E(0,-2)$, $F(2,-4)$, $G(4,-2)$, $H(2,0)$).

Step3: Calculate new coordinates of E

$x_E'=0 - 4=-4$, $y_E'=-2 + 6 = 4$. So, $E'(-4,4)$.

Step4: Calculate new coordinates of F

$x_F'=2 - 4=-2$, $y_F'=-4 + 6 = 2$. So, $F'(-2,2)$.

Step5: Calculate new coordinates of G

$x_G'=4 - 4 = 0$, $y_G'=-2+6 = 4$. So, $G'(0,4)$.

Step6: Calculate new coordinates of H

$x_H'=2 - 4=-2$, $y_H'=0 + 6 = 6$. So, $H'(-2,6)$.

Answer:

The coordinates of the image $E'F'G'H'$ are $E'(-4,4)$, $F'(-2,2)$, $G'(0,4)$, $H'(-2,6)$