The vertices of JKL are J(4,3), K(-3,4), L(-1,-1). Graph the image of JKL after a reflection in the line y=-x.

y=-x

J(-4,3)

K(3,-4)

L(1,1)

where's the diagrams?

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.