Question

graph the image of trapezoid bcde after a translation 4 units left.

Explanation:

Step1: Recall translation rule

For a translation 4 units left, subtract 4 from the x - coordinates of each vertex.

Step2: Identify original vertices

Assume \(B(-6,-8)\), \(C(-6,1)\), \(D(1,0)\), \(E(1,-8)\).

Step3: Calculate new vertices

For \(B\): New \(x=-6 - 4=-10\), \(y = - 8\), so \(B'(-10,-8)\).
For \(C\): New \(x=-6 - 4=-10\), \(y = 1\), so \(C'(-10,1)\).
For \(D\): New \(x=1 - 4=-3\), \(y = 0\), so \(D'(-3,0)\).
For \(E\): New \(x=1 - 4=-3\), \(y=-8\), so \(E'(-3,-8)\).

Step4: Graph new trapezoid

Plot \(B'(-10,-8)\), \(C'(-10,1)\), \(D'(-3,0)\), \(E'(-3,-8)\) and connect them to form the new trapezoid.

Answer:

Graph the trapezoid with vertices \(B'(-10,-8)\), \(C'(-10,1)\), \(D'(-3,0)\), \(E'(-3,-8)\) on the given coordinate - plane.