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Ber [7]
3 years ago
12

Using the completing-the-square method, rewrite f(x) = x2 - 6x + 2 in vertex form.​

Mathematics
1 answer:
sp2606 [1]3 years ago
8 0

Answer:

\large\boxed{f(x)+(x-3)^2-7}

Step-by-step explanation:

\text{The vertex form of equation}\ y=ax^2+bx+c:\\\\y=a(x-h)^2+k\\\\\text{We have the equation:}\\\\f(x)=x^2-6x+2=x^2-2(x)(3)+1=x^2-2(x)(3)+3^2-3^2+2\\\\\text{Use}\ (a-b)^2=a^2-2ab+b^2\to x^2-2(x)(3)+3^2=(x-3)^2\\\\f(x)=(x-3)^2-9+2=(x-3)^2-7

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PLSSS ANSWER, 35 pointss
xxTIMURxx [149]

Answer:

1.

(y - 5)(x + y)

2.

{(x - y)}(x - y   + a)

3.

(b - 3)(5b - 17)

Step-by-step explanation:

1. The given polynomial is

x(y - 5) - y(5 - y)

We need to factor the second part to get:

x(y - 5)  +  y(y - 5)

We factor y-5 to obtain:

(y - 5)(x + y)

2. We have

{(x - y)}^{2}  - a(y - x)

We again factor -1 from the second part to get:

{(x - y)}^{2} +  a(x- y)

We now factor x-y to get:

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3. We have

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We factor negative 1 from the first part to get

- 2(b - 3) + 5 {(b - 3})^{2}

We now factor b-3 to get:

(b - 3)( - 2 + 5(b - 3))

We now simplify to get:

(b - 3)( - 2 + 5b - 15) = (b - 3)(5b - 17)

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Step-by-step explanation:

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