Do the graphs of 4y=(x-3)^2 and 3x-y=14 intersect at (5,1)
1 answer:
Answer:
it will happen if both y and x components be equal
so
y= (x-3)²/4 , y = 3x- 14
3x - 14 = (x-3)²/4
12x - 56 = x²-6x + 9
x² - 18x + 65 = 0
delta = 18²-4*65 = 64
x1 = (18 + (64)^(1/2))/2 = 13
x2 = (18 - (64)^(1/2))/2 = 5
one point is x = 5 but we should check if on that point y component is 1 or not
y = (5-3)²/4 = 1
so the functions will intersect at (5,1)
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