△def is reflected to form △d′e′f′ . the vertices of △def are d(−1, −1) , e(1, −1) , and f(−1, −6) . the vertices of △p′q′r′ are d′(−1, 1) , e′(1, 1) , and f′(−1, 6) . which reflection results in the transformation of△defto△d′e′f′ ? reflection across the x-axis reflection across the y-axis reflection across y = x reflection across y=−x

To find the answer first graph DEF and D'E'F'. Then

c

step-by-step explanation:

c

step-by-step explanation:

there is vertical asymptote at x = 2 and x = 0

vertical asymptote occurs at the points where the denominator is 0. so we can say that

x = 2 occurs when there is a factor (x-2) in the denominator, and

x = 0 occurs when there is a factor "x" in the denominator

thus, the denominator of the equation would have "x * (x-2)", this is option c only. that's the correct answer.