Given: EGEF = JK Prove: FG = HJ PLEASE HELP ASP
1 answer:
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Answer:
Step-by-step explanation:
i) 1
ii) see attached
iii) as x goes from 0 to 2, y decreases parabolically from 1 to an acme at -3
iv) y = (x - 2)² - 3
v) -3 < t < 1
if t < -3 no intersection occurs between y = t and y = (x - 2)² - 3
if t = -3, y = t intersects the curve at only one point, (2, -3)
if -3 < t < 1 there exist two intersecting points with positive x values
if t = 1 there exist only one positive x value intersection as the other is at (0, 1) and 0 is neither positive nor negative.
if t > 1 there is one positive x value intersection and one negative x value intersection.
