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defon
1 year ago
9

What is a solution to the system of equations that includes quadratic function f(x) and linear function g(x)? f(x) = 2x2 + x + 4

x g(x) –2 1 –1 3 0 5 1 7 2 9
Mathematics
1 answer:
olga_2 [115]1 year ago
5 0

The solution to the system of equations that includes quadratic function f(x) and linear function g(x) is

(\sqrt{2},\sqrt{-2}  ) and ((7+\sqrt{2)} ,(7+\sqrt{-2}))

We have given that,

f(x) = 2x^2 + x + 4

x                          g(x)      

-2                               1      

  -1                              3        

  0                              5      

  1                              7

  2                              9

<h3>What is a polynomial function?</h3>

A polynomial function is a relation where a dependent variable is equal to a  polynomial expression.

A polynomial expression is an expression including numbers and variables, where variables are raised to non-negative powers.

The general form of a polynomial expression is:

a₀ + a₁x + a₂x² + a₃x³ + ... + anxⁿ.

The highest power to a variable is the degree of the polynomial expression. When degree = 2, the function is a quadratic function.

When degree = 1, the function is a linear function.

<h3>How do we solve the given question?</h3>

The quadratic function is given to us:f(x) = 2x^2 + x + 4.

We need to determine the linear equation g(x).

Since it's a linear equation we use the two-point method to determine the equation.

<h3>What is the two-point?</h3>

y-y₁ = ((y₂-y₁)/(x₂-x₁))*(x-x₁)

We take the points g(-2) =1, g(-1) = 3

g(x) - g(1) = ((g(-2)-g(-1))/(-2+1))*(x-1)or,

g(x) - 1 = ((-2-(-1))/(-2+1))*(x-1)or, g(x) - 1 = -1(-x+1)or,

g(x) = x - 1 + 1 = x

∴ g(x) = x , is the linear function g(x)

We are asked to find the solution to the system of equations f(x) and g(x).To find the solution we need to check what is the common solution to both f(x) and g(x).

For that, we equate f(x) and g(x).2x^2 + x + 4 = x or, 2x² - x +x- 4 = 0or, 2x^2  - 4 = 02(x^2-2)=0x^2-2=0x^2-\sqrt{2}=0(x-\sqrt{2}) (x+\sqrt{2})=0(x-\sqrt{2})=0 \\x=\sqrt{2}  or x=-\sqrt{2}g(-1) = 3(from the table)g(\sqrt{2})=\sqrt{2}  \ and \ g(\sqrt{-2}) =-\sqrt{2}f(\sqrt{2}) = 2(\sqrt{2} )^2 + (\sqrt{2} ) + 3\\=2(2)+\sqrt{2} +3\\=7+\sqrt{2}g(-\sqrt{2} )= 2(\sqrt{-2} )^2 + (\sqrt{-2} ) + 3\\=2(2)+\sqrt{-2} +3\\=7+\sqrt{-2}

The solution to the system of equations that includes quadratic function f(x) and linear function g(x) is (\sqrt{2},\sqrt{-2}  ) and ((7+\sqrt{2)} ,(7+\sqrt{-2}))

Learn more about linear and  quadratic equations at

brainly.com/question/14075672

#SPJ1

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Case 2: As x > 0 increases, f(x) decreases.  As x < 0 decreases, f(x) decreases.

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Step-by-step explanation:

Let us consider a monomial function:

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Case 1:

As x > 0 increases, f(x) increases.  As x < 0 decreases, f(x) decreases.

This happens only if a is 'positive' and n is 'odd'.  So, it is an 'odd' function with 'positive' a.

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This happens only if a is 'negative' and n is 'odd'. So, it is an 'odd' function with 'negative' a.

Keywords: monomial function, odd function, even function

Learn more about monomial function from brainly.com/question/8973176

#learnwithBrainly

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