#Inequalities using graphs.

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spare eagleBOT
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shut plinth
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If the shaded part lies above the graph/points then its y>, otherwise if the shaded part lies below the graph/points then its y<

olive helm
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You pick one point to test. Say, for this parabola, you pick the point (0,0). $(x^2-6x+5)\vert_{x=0} = 5$, which means the point (0,0) is BELOW the graph of the function because $0 < 5$.
Are you asking why you only need to test one point?

plush cometBOT
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gfauxpas

205675981845954560.png
broken tinsel
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but how does that logic apply for example for circles?

shut plinth
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well a circle is defined implicitly not explicitly anyway so youd need to apply a different type of logic
(a circle isnt defined as y =... , rather its defined as x^2 + y^2 = r^2)

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if x^2 + y^2 < r^2 that means that we're considering all points a distance LESS than r from the origin, so youd shade the inside of the circle

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If x^2 + y^2 > r^2 that means that we're considering all points a distance MORE than r from the origin, so you shade everything outside of the circle

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ie this is x^2+y^2 > 5^2

image.png
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this is x^2 + y^2 < 5^2

image.png
shut cloak
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simple

olive helm