Question

graph the image of δtuv after a reflection over the x-axis.

Explanation:

Step1: Find coordinates of T, U, V

First, we identify the coordinates of the vertices of \(\triangle TUV\). From the graph:

• Point \(T\): Let's assume the grid has each square as 1 unit. Looking at the graph, \(T\) is at \((-4, -6)\) (since it's 4 units left on the x - axis and 6 units down on the y - axis).
• Point \(U\): \(U\) is at \((4, -6)\) (4 units right on the x - axis and 6 units down on the y - axis).
• Point \(V\): \(V\) is at \((-3, -8)\) (3 units left on the x - axis and 8 units down on the y - axis).

Step2: Apply reflection over x - axis rule

The rule for reflecting a point \((x,y)\) over the \(x\) - axis is \((x,y)\to(x, -y)\).

• For point \(T(-4, -6)\): Applying the rule, the reflected point \(T'\) will be \((-4, -(-6))=(-4,6)\).
• For point \(U(4, -6)\): Applying the rule, the reflected point \(U'\) will be \((4, -(-6))=(4,6)\).
• For point \(V(-3, -8)\): Applying the rule, the reflected point \(V'\) will be \((-3, -(-8))=(-3,8)\).

Step3: Plot the reflected points

Now, we plot the points \(T'(-4,6)\), \(U'(4,6)\), and \(V'(-3,8)\) on the coordinate plane and connect them to form the reflected triangle \(\triangle T'U'V'\).

Answer:

The image of \(\triangle TUV\) after reflection over the \(x\) - axis has vertices at \(T'(-4,6)\), \(U'(4,6)\), and \(V'(-3,8)\). (To graph it, plot these points and draw the triangle connecting them.)