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y<8^2 => y<64
This is your y "limit". You can draw a horizontal line on the graph that goes through the y coordinate 64, then shade the area below it.
Now we have to look at the second inequality. There are two squared terms so this is likely just a circle. We have to work a bit on the expression to be sure:
9x^2 - 36x + 4y^2 - 16y <-16
You can see that all the terms are perfect squares so this suggests we can group them using completion of the square
(9x^2 - 36x + 36) + (4y^2 - 16y + 16) < -16 +36 +16
=> (3x-6)^2 + (2y-4)^2 < 36
So it is indeed a circle. Its center is at (6,4) and has a radius of 6. Draw this circle and shade the inside area.
The area that is shaded by both the first and the second inequality represent the solution(in other words, the area inside the circle).
This site needs an edit feature. Right when completing the square:
9(x^2-4x)+4(y^2-4y) < -16
9(x^2-4x+4)+4(y^2-4y+4) < 36
divide by 36 and rewrite the paranthesis
[(x-2)^2]/4 + [(y-2)^2]/9 < 1
The expression is actually an ellipse and has the center at (2,2), the x radius = 2 and y radius = 3.
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