Question

graph the image of △tuv after a dilation with a scale factor of 3, centered at the origin.

Explanation:

Step1: Identify the coordinates of vertices

Let's assume the coordinates of the vertices of $\triangle TUV$ are $T(-3, - 1)$, $U(3, - 1)$, $V(-3,2)$.

Step2: Apply the dilation formula

For a dilation centered at the origin with scale - factor $k = 3$, the formula to find the coordinates of the dilated point $(x',y')$ from the original point $(x,y)$ is $(x',y')=(k\cdot x,k\cdot y)$.
For point $T(-3,-1)$:
$x_T'=3\times(-3)=-9$, $y_T'=3\times(-1)=-3$. So $T'(-9,-3)$.
For point $U(3,-1)$:
$x_U'=3\times3 = 9$, $y_U'=3\times(-1)=-3$. So $U'(9,-3)$.
For point $V(-3,2)$:
$x_V'=3\times(-3)=-9$, $y_V'=3\times2 = 6$. So $V'(-9,6)$.

Step3: Graph the new triangle

Plot the points $T'(-9,-3)$, $U'(9,-3)$ and $V'(-9,6)$ on the coordinate - plane and connect them to form $\triangle T'U'V'$.

Answer:

Graph the triangle with vertices $T'(-9,-3)$, $U'(9,-3)$ and $V'(-9,6)$ on the given coordinate - grid.