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Factored form of f(x)=-(x-7)^2+1

User Deadfishli
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Answer:

Hope you understand!

Explanation:

The factored form of the function f(x) = -(x-7)^2 + 1 can be determined by expanding the square and simplifying.

Let's start by expanding the square:

f(x) = -(x-7)(x-7) + 1

To simplify, we can use the FOIL method:

f(x) = -(x^2 - 7x - 7x + 49) + 1

f(x) = -(x^2 - 14x + 49) + 1

f(x) = -x^2 + 14x - 49 + 1

f(x) = -x^2 + 14x - 48

So, the factored form of f(x) = -(x-7)^2 + 1 is -x^2 + 14x - 48.

In the factored form, the polynomial is expressed as a product of factors. However, in this case, the given function is not in factored form. It is in standard form, which is in the form of ax^2 + bx + c, where a, b, and c are constants.

To convert the function to factored form, we need to complete the square. This process involves rearranging the terms to create a perfect square trinomial.

Starting with the standard form -x^2 + 14x - 48, we can complete the square as follows:

1. Group the x terms and keep the constant term separate:

-x^2 + 14x - 48 = -(x^2 - 14x) - 48

2. To create a perfect square trinomial, take half of the coefficient of x (-14) and square it, which gives us 49:

-(x^2 - 14x + 49) - 48

3. Simplify and combine like terms:

-(x^2 - 14x + 49) - 48 = -(x - 7)^2 - 48

So, the factored form of f(x) = -(x-7)^2 + 1 is -(x - 7)^2 - 48.

User Monzur
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