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1 answer:
Given:
The system of equations is
To find:
The solution of given system of equations.
Solution:
we have,
...(i)
...(ii)
From (i) and (ii), we get
Divide both sides by 3.
Putting x=2 in (i), we get
So, the solution is (2,4).
For ,
x y
0 0
1 2
For ,
x y
0 6
1 5
Plot these points on a coordinate plane and connect them by straight line as shown below.
From the below graph it is clear that both lines intersect each other at (2,4). So, the solution is (2,4).
Therefore, the solution of system of equations is (2,4).
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Answer:
- zeros are {-2, 3, 7} as verified by graphing
- end behavior: f(x) tends toward infinity with the same sign as x
Step-by-step explanation:
A graphing calculator makes finding or verifying the zeros of a polynomial function as simple as typing the function into the input box.
<h3>Zeros</h3>The attachment shows the function zeros to be x ∈ {-2, 3, 7}, as required.
<h3>End behavior</h3>The leading coefficient of this odd-degree polynomial is positive, so the value of f(x) tends toward infinity of the same sign as x when the magnitude of x tends toward infinity.
- x → -∞; f(x) → -∞
- x → ∞; f(x) → ∞
__
<em>Additional comment</em>
The function is entered in the graphing calculator input box in "Horner form," which is also a convenient form for hand-evaluation of the function.
We know the x^2 coefficient is the opposite of the sum of the zeros:
-(7 +(-2) +3) = -8 . . . . x^2 coefficient
And we know the constant is the opposite of the product of the zeros:
-(7)(-2)(3) = 42 . . . . . constant
These checks lend further confidence that the zeros are those given.
(The constant is the opposite of the product of zeros only for odd-degree polynomials. For even-degree polynomials. the constant is the product of zeros.)
