Write a third degree polynomial function that has real zeros -1, -2 and 4

Mathematics

1 answer:

ra1l [238]3 years ago

8 0

Answer:

f(x) = $x^{3} - x^{2} - 10x -8$

Step-by-step explanation:

All you have to do is multiply (x + 1)(x + 2)(x - 4)

You should get $x^{3} - x^{2} - 10x -8$

So, f(x) = $x^{3} - x^{2} - 10x -8$

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3 years ago

Graph the system of inequalities presented here on your own paper, then use your graph to answer the following questions: y <

levacccp [35]

Solution-

The given inequalities are,

$y < 2x-7,\\\\y\geq-\frac{1}{2}x+3$

Part A

As the former inequality is only '<' so it will be represented as dotted line, but in case of the later inequality there is '≥' so it will be represented as a solid line.

This happens because in only '<, >' the values must be either lesser or greater than y, they can't take the values of y, but in case of '≤, ≥' the values can be equal to y.

Taking origin (0, 0) and putting in the first inequality.

$\Rightarrow 0< 2(0)-7\\\\\Rightarrow 0\nless -7$

As the equation does not satisfy, so the shaded area will be on the other side of the origin.

Again taking origin (0, 0) and putting in the second inequality.

$\Rightarrow 0\geq-\frac{1}{2}0+3\\\\\Rightarrow 0\ngeqslant 3$

As the equation does not satisfy, so the shaded area will be on the other side of the origin.

The second attachment contains the feasible region.

Part B

In order to know whether the point (3, −7) is included in the solution area for the system or not, we must put it in both the equations, if it satisfies both then only it is within the solution area.

$\Rightarrow -7< 2(3)-7,\ -7\geq-\frac{1}{2}(3)+3\\\\\Rightarrow -7$