Two quantities, x and y, are directly proportional. If you take a third of x, what happens to y
2 answers:
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To solve this we just need a polynomial where the roots can be -10 so in (x -/+ N)
The Ns must equal -10
We also know there must be at least a degree of 2 or higher, so we want X^3 or 3 roots. Given this we can construct our function;
H(x) = (X-2)(X+5)(X+1)
1*5*-2 = -10
So multiplying that out to get the standard form
X^2-2X+5X-10(X+1)
Simplifying to X^2 +3X-10(X+1)
X^3+3X^2-10X + X^2 +3X -10
Which simplifies to:
X^3+4X^2-7X-10
And below the desmos shows the y-int at (0,-10)
The Ns must equal -10
We also know there must be at least a degree of 2 or higher, so we want X^3 or 3 roots. Given this we can construct our function;
H(x) = (X-2)(X+5)(X+1)
1*5*-2 = -10
So multiplying that out to get the standard form
X^2-2X+5X-10(X+1)
Simplifying to X^2 +3X-10(X+1)
X^3+3X^2-10X + X^2 +3X -10
Which simplifies to:
X^3+4X^2-7X-10
And below the desmos shows the y-int at (0,-10)