does point (1,-2) lie on the graph of the polynomial -3x+2x+1
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Points-(1,-2)
Hence, x=1 and y=-2
Putting values in the equation
-3x+2x+1 = 0
-3(1) +2(-2) +1 = 0
-3 -4 +1 =0
-6=0
as LHS is not equal to RHS
thus implies the point(1,-2) does not lie on the graph of the polynomial -3x+2x+1
Hence, x=1 and y=-2
Putting values in the equation
-3x+2x+1 = 0
-3(1) +2(-2) +1 = 0
-3 -4 +1 =0
-6=0
as LHS is not equal to RHS
thus implies the point(1,-2) does not lie on the graph of the polynomial -3x+2x+1
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Brainly Verified
No, (1,-2) don't lies on the graph of p(x)
some users might get confused that I changes p(x) into y.
See,
if this was an expression with no 'y' in it then I would have added a 0y in it, but this is an polynomial(equation) so in these kind of cases we changes p(x) into y
And if this equation was having the variable y then I would have changed p(y) into x.
Hope this ans helped you.
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