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3 years ago

Is y= -2 a function? Someone please help me ✌️✨

Mathematics

2 answers:

o-na [289]3 years ago

5 0

Answer:

Step-by-step explanation:

A function is defined as a set of outputs based on a set of inputs.

This means that we need to have two variables in order for this to be a function.

y needs to change as we input different values of x.

However, there are no x values here! It's just a simple line.

Therefore this is not a function.

Hope this helped!

lisov135 [29]3 years ago

3 0

Answer:

no it's not

hope it helps;)

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The roots of f(x) are: -2, 3, (1+3i) and (1-3i)

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(1+3i) is a root of f(x)

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Since, (1+3i) is a root of f(x),

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Now, we check if (1 - 3i) is a root of given function.

$f(x)=x^4-3x^3+6x^2+2x-60\\f(1-3i)=(1-3i)^4-3(1-3i)^3+6(1-3i)^2+2(1-3i)-60\\= (28+96i)-3(-26+18i)+6(-8-6i)+2(1-3i)-60 = 0$

Thus, (1-3i) is also a root of given function.

Since, (1-3i) is a root of f(x),

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is a factor of given expression.

Thus, we can write:

$(x-(1+3i))(x-(1-3i))\text{ is a factor of f(x)}\\x^2-x(1-3i)+x(1+3i)+(1-3i)(1+3i)\text{ is a factor of f(x)}\\x^2-2x+10\text{ is a factor of f(x)}$

Dividing f(x) by above expression:

$\displaystyle\frac{x^4-3x^3+6x^2+2x-60}{x^2-2x+10} = x^2-x-6$

To find the root, we equate it to zero:

$x^2-x-6 = 0\\x^2 - 3x+2x-6=0\\x(x-3)+2(x-3)=0\\(x+2)(x-3) = 0\\x = -2, x = 3$

Thus, the roots of f(x) are: -2, 3, (1+3i) and (1-3i)

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