Question

graph the image of △hij after the following sequence of transformations: reflection across the line x = -1 translation 6 units left and 18 units up

Explanation:

Step1: Recall reflection formula

For a point $(x,y)$ reflected across the line $x = a$, the new $x$-coordinate is $2a - x$ and $y$-coordinate remains the same. Here $a=-1$. Let the vertices of $\triangle HIJ$ be $(x_1,y_1),(x_2,y_2),(x_3,y_3)$. After reflection across $x = - 1$, the new vertices are $(2\times(-1)-x_1,y_1),(2\times(-1)-x_2,y_2),(2\times(-1)-x_3,y_3)$.

Step2: Recall translation formula

A translation of $h$ units left and $k$ units up for a point $(x,y)$ gives a new point $(x - h,y + k)$. Here $h = 6$ and $k=18$. So for the reflected points from Step 1, the final - transformed points are $(2\times(-1)-x_1-6,y_1 + 18),(2\times(-1)-x_2-6,y_2 + 18),(2\times(-1)-x_3-6,y_3 + 18)$.

Step3: Graph the new triangle

Plot the new - transformed vertices on the coordinate plane and connect them to form the image of $\triangle HIJ$.

Since we don't have the coordinates of the vertices of $\triangle HIJ$, we can't give numerical answers. But the general procedure to graph the image is as above.

Answer:

Graph the triangle with vertices obtained by first reflecting across $x=-1$ and then translating 6 units left and 18 units up.