Question

graph the image of trapezoid cdef after a dilation with a scale factor of $\frac{1}{4}$, centered at the origin.

Explanation:

Step1: Identify trapezoid vertices

Assume \(C(-8,-4)\), \(D(8,-4)\), \(E(6,8)\), \(F(-8,8)\)

Step2: Apply dilation formula

For a dilation centered at the origin with scale - factor \(k = \frac{1}{4}\), the formula for a point \((x,y)\) is \((x',y')=(k x,k y)\)
For point \(C\): \(x'_C=\frac{1}{4}\times(-8)= - 2\), \(y'_C=\frac{1}{4}\times(-4)=-1\)
For point \(D\): \(x'_D=\frac{1}{4}\times8 = 2\), \(y'_D=\frac{1}{4}\times(-4)=-1\)
For point \(E\): \(x'_E=\frac{1}{4}\times6=\frac{3}{2}\), \(y'_E=\frac{1}{4}\times8 = 2\)
For point \(F\): \(x'_F=\frac{1}{4}\times(-8)=-2\), \(y'_F=\frac{1}{4}\times8 = 2\)

Step3: Graph new trapezoid

Plot the new points \(C'(-2,-1)\), \(D'(2,-1)\), \(E'(\frac{3}{2},2)\), \(F'(-2,2)\) and connect them to form the dilated trapezoid.

Answer:

Graph the trapezoid with vertices \(C'(-2,-1)\), \(D'(2,-1)\), \(E'(\frac{3}{2},2)\), \(F'(-2,2)\)