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Leno4ka [110]
3 years ago
9

Write the function in vertex form. y = 2x² + x

Mathematics
2 answers:
Sliva [168]3 years ago
6 0
<h2>Steps:</h2>
  • Vertex Form (aka general form): y=a(x-h)^2+k , with (h,k) as the vertex.

So for this, we are going to be completing the square. Firstly, we have to factor out the 2 out of the right side of the equation to make the x² coefficient 1:

y=2(x^2+\frac{1}{2}x)

Next, we want to make the quantity inside of the parentheses a perfect square. To find the constant of this soon-to-be perfect square, divide the x coefficient by 2 and then square the quotient. In this case:

\frac{1}{2}\div \frac{2}{1}\\\\\frac{1}{2}\times\frac{1}{2}=\frac{1}{4}\\\\\\(\frac{1}{4})^2=\frac{1^2}{4^2}=\frac{1}{16}

Now, we are going to be adding 1/16 inside of the parentheses. Now we want to cancel this quantity out (note: addition property of equality). To do this, we want to add the product of 1/16 and 2 (since 2 is multiplying with 1/16 on the right side) to the left side:

\frac{1}{16}\times\frac{2}{1}=\frac{2}{16}=\frac{1}{8}\\\\y+\frac{1}{8}=2(x^2+\frac{1}{2}x+\frac{1}{16})

Now that the quantity inside the parentheses is a perfect square, factor:

  • Tip: (x+y)^2=x^2+2xy+y^2

x^2+\frac{1}{2}x+\frac{1}{16}=(x+\frac{1}{4})^2\\\\y+\frac{1}{8}=2(x+\frac{1}{4})^2

Lastly, subtract both sides by 1/8:

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

<h2>Answer:</h2>

<u>In short, your vertex form is: y=2(x+\frac{1}{4})^2-\frac{1}{8}</u>

N76 [4]3 years ago
4 0

To start to solve this problem, we need to know what vertex form is. The vertex form of a parabola is. The vertex form of a parabola is a(x-h) + k, where k is the vertical shift, h is the horizontal shift, and a is the value that tells the stretch.


To start to solve this equation, we want to start to create a difference of two squares.

y = 2(x²+\frac{1}{2}x) We do this step to make the x² have a coefficient of 1

Now, we want to complete the square. To complete the square, we take 1/2 of the coefficient of x, and then square that.

1/2 * 1/2 = 1/4, and 1/4²=1/16

That means that we need to add 1/16 inside and outside the parenthesis.

We get:

y = 2(x²+1/2x + 1/16) - 1/16*2

We do -1/16*2 on the outside because since we added it inside the parenthesis, we need to take it away somewhere else (if that makes sense). The two is there because there is a two in front of the parenthesis.

We get:

y = 2(x+1/4)² - 1/8, by completing the square and simplifying, and this is the final answer.



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A total of 12 players consisting 6 male and 6 female badminton players are attending a training camp
abruzzese [7]

Step-by-step explanation:

<em>"A total of 12 players consisting 6 male and 6 female badminton players are attending a training camp."</em>

<em />

<em>"(a) During a morning activity of the camp, these 12 players have to randomly group into six pairs of two players each."</em>

<em>"(i) Find the total number of possible ways that these six pairs can be formed."</em>

The order doesn't matter (AB is the same as BA), so use combinations.

For the first pair, there are ₁₂C₂ ways to choose 2 people from 12.

For the second pair, there are ₁₀C₂ ways to choose 2 people from 10.

So on and so forth.  The total number of combinations is:

₁₂C₂ × ₁₀C₂ × ₈C₂ × ₆C₂ × ₄C₂ × ₂C₂

= 66 × 45 × 28 × 15 × 6 × 1

= 7,484,400

<em>"(ii) Find the probability that each pair contains players of the same gender only. Correct your final answer to 4 decimal places."</em>

We need to find the number of ways that 6 boys can be grouped into 3 pairs.  Using the same logic as before:

₆C₂ × ₄C₂ × ₂C₂

= 15 × 6 × 1

= 90

There are 90 ways that 6 boys can be grouped into 3 pairs, which means there's also 90 ways that 6 girls can be grouped into 3 pairs.  So the probability is:

90 × 90 / 7,484,400

= 1 / 924

≈ 0.0011

<em>"(b) During an afternoon activity of the camp, 6 players are randomly selected and 6 one-on-one matches with the coach are to be scheduled.</em>

<em>(i) How many different schedules are possible?"</em>

There are ₁₂C₆ ways that 6 players can be selected from 12.  From there, each possible schedule has a different order of players, so we need to use permutations.

There are 6 options for the first match.  After that, there are 5 options for the second match.  Then 4 options for the third match.  So on and so forth.  So the number of permutations is 6!.

The total number of possible schedules is:

₁₂C₆ × 6!

= 924 × 720

= 665,280

<em>"(ii) Find the probability that the number of selected male players is higher than that of female players given that at most 4 females were selected. Correct your final answer to 4 decimal places."</em>

If at most 4 girls are selected, that means there's either 0, 1, 2, 3, or 4 girls.

If 0 girls are selected, the number of combinations is:

₆C₆ × ₆C₀ = 1 × 1 = 1

If 1 girl is selected, the number of combinations is:

₆C₅ × ₆C₁ = 6 × 6 = 36

If 2 girls are selected, the number of combinations is:

₆C₄ × ₆C₂ = 15 × 15 = 225

If 3 girls are selected, the number of combinations is:

₆C₃ × ₆C₃ = 20 × 20 = 400

If 4 girls are selected, the number of combinations is:

₆C₂ × ₆C₄ = 15 × 15 = 225

The probability that there are more boys than girls is:

(1 + 36 + 225) / (1 + 36 + 225 + 400 + 225)

= 262 / 887

≈ 0.2954

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Step-by-step explanation:

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Factor the numerator:

(q + 7) (q − 3) / (q − 3)

Divide:

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