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murzikaleks [220]
3 years ago
12

identify the vertex, axis of symmetry, maximum or minimum for each of the following quadratic equations​

Mathematics
2 answers:
saul85 [17]3 years ago
5 0

check it please and tell me

olga_2 [115]3 years ago
3 0

Answer:  see below

<u>Step-by-step explanation:</u>

The vertex form of a quadratic equation is:    y = a(x - h)² + k     where

  • "a" is the vertical stretch (positive = min [U], negative = max [∩])
  • (h, k) is the vertex
  • Axis of Symmetry is always: x = h
  • Domain is always: x = All Real Numbers
  • Range is y ≥ k when "a" is positive or y ≤ k when "a" is negative

6) y = 3(x - 1)² + 0

       ↓       ↓      ↓

     a= +   h= 1   k= 0

Vertex: (h, k) = (1, 0)

Axis of Symmetry: x = h  →  x = 1  

Max/Min: "a" is positive → minimum

Domain: x = All Real Numbers

Range:  y ≥ k → y ≥ 0

7) y = -1/2(x + 6)² - 7

          ↓       ↓      ↓

        a= -   h= -6  k= -7

Vertex: (h, k) = (-6, -7)

Axis of Symmetry: x = h  →  x = -6  

Max/Min: "a" is negative → maximum

Domain: x = All Real Numbers

Range:  y ≤ k → y ≤ -7

8) y = -(x - 0)² - 3

       ↓       ↓      ↓

     a= -   h= 0   k= -3

Vertex: (h, k) = (0, -3)

Axis of Symmetry: x = h  →  x = 0  

Max/Min: "a" is negative → maximum

Domain: x = All Real Numbers

Range:  y ≤ k → y ≤ -3

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Riley makes a mistake in step 2 while doing her homework. What was her mistake? StartFraction x Over x squared minus 5 x + 6 End
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Answer:

The answer is:

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Step-by-step explanation:

Solving the given equation

\frac{x}{x^2-5x+6}+\frac{x}{x+3}\\\\Step 1: \frac{x}{(x-2)(x-3)}+\frac{x}{x+3}\\Step2: \frac{x}{(x-2)(x-3)}+\frac{x(x-2)}{(x-2)(x+3)}\\Step3:\frac{x(x+3)+x(x-2)(x+3)}{(x-2)(x-3)(x+3)}

Lets consider the steps given in the question:

Step2: \frac{x}{(x-2)(x+3)}+\frac{3(x-2)}{(x-2)(x+3)}

Step3: \frac{x+3(x-2)}{(x-2)(x+3)}\\

Lets compare both solutions:

In the original solution, the denominator of first term is (x-2)(x-3)

In the solution given in the question, the denominator of first term is (x-2)(x+3).

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