The product of this problem

Marsha wants to determine the vertex of the quadratic function f(x) = x2 – x + 2. What is the function’s vertex?

BaLLatris [955]

Answer:

$Vertex = (\frac{1}{2},\frac{7}{4})$

Step-by-step explanation:

Given

$f(x) = x^2 - x +2$

Required

The vertex

We have:

$f(x) = x^2 - x +2$

First, we express the equation as:

$f(x) = a(x - h)^2 +k$

Where

$Vertex = (h,k)$

So, we have:

$f(x) = x^2 - x +2$

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Take the coefficient of x: -1

Divide by 2: (-1/2)

Square: (-1/2)^2

Add and subtract this to the equation

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$f(x) = x^2 - x +2$

$f(x) = x^2 - x + (-\frac{1}{2})^2+2 -(-\frac{1}{2})^2$

$f(x) = x^2 - x + \frac{1}{4}+2 -\frac{1}{4}$

Expand

$f(x) = x^2 - \frac{1}{2}x- \frac{1}{2}x + \frac{1}{4}+2 -\frac{1}{4}$

Factorize

$f(x) = x(x - \frac{1}{2})- \frac{1}{2}(x - \frac{1}{2})+2 -\frac{1}{4}$

Factor out x - 1/2

$f(x) = (x - \frac{1}{2})(x - \frac{1}{2})+2 -\frac{1}{4}$

$f(x) = (x - \frac{1}{2})^2+2 -\frac{1}{4}$

$f(x) = (x - \frac{1}{2})^2+ \frac{8 -1 }{4}$

$f(x) = (x - \frac{1}{2})^2+ \frac{7}{4}$

Compare to: $f(x) = a(x - h)^2 +k$

$h = \frac{1}{2}$

$k = \frac{7}{4}$

Hence:

$Vertex = (\frac{1}{2},\frac{7}{4})$

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How to find the area

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L + W
length plus width

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Rudiy27

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Step-by-step explanation:

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Help. please read the instructions and Answer the questions

Dominik [7]

9514 1404 393

Explanation:

We refer to the equations as [1] and [2]. We refer to the items as (1) – (4).

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1. The terms to be eliminated have matching coefficients in (1) and (2). They can be eliminated by subtracting one equation from the other.

In (3) and (4), putting the equations in standard form* results in terms with opposite coefficients. Those terms can be eliminated by adding the equations.

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2. Terms to be eliminated will have matching or opposite coefficients.

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3. In (1) and (2), the variable x can be eliminated by subtracting one equation from the other. In the attachment, we have indicated the subtraction that will result in the remaining variable having a positive coefficient.

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4. In (3) and (4), the coefficients of the variables are not equal or opposite in the two equations, so no variable can be eliminated directly.

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5. As suggested by the answer to Q4, an equivalent equation must be found that has an equal or opposite variable coefficient with respect to the other equation. The new equations are ...

(3) [2] ⇒ x -y = 2

(4) [1] ⇒ 2x +2y = 3

_____

Here are the solutions:

(1) [1] -[2] ⇒ (x +y) -(x -y) = (-1) -(3)

2y = -4 ⇒ y = -2

x = y +3 = 1 . . . . from [2]

(x, y) = (1, -2)

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(2) [2] -[1] ⇒ (x +2y) -(x +y) = (8) -(5)

y = 3

x = 5 -y = 2 . . . . from [1]

(x, y) = (2, 3)

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(3) [1] +[2]/2 ⇒ (x +y) +(x -y) = (1) +(2)

2x = 3 ⇒ x = 3/2

y = 1 -x = -1/2 . . . . from [1]

(x, y) = (3/2, -1/2)

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(4) [2] +[1]/2 ⇒ (5x -2y) +(2x +2y) = (4) +(3)

7x = 7 ⇒ x = 1

y = (3 -2x)/2 = 1/2

(x, y) = (1, 1/2)

_____

* Equations in standard form have mutually prime coefficients. In (3) a factor of 2 can be removed from equation [2]. In (4), a factor of 2 can be removed from equation [1].

7 0

3 years ago