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Rewrite standard form y=3x^2 + 12x + 14 into vertex form

User Zzarbi
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2 Answers

7 votes

Answer:

the vertex form of the quadratic equation y=3x^2 + 12x + 14 is y = 3(x + 2)^2 - 2. The vertex of this parabola is located at the point (-2, -2).

Explanation:

To rewrite the quadratic equation y=3x^2 + 12x + 14 into vertex form, we need to complete the square.

First, we can factor out the coefficient of x^2, which is 3:

y = 3(x^2 + 4x) + 14

Next, we need to add and subtract a constant term inside the parentheses that will allow us to complete the square. We can do this by adding and subtracting (4/2)^2 = 4:

y = 3(x^2 + 4x + 4 - 4) + 14

Now we can write the first three terms inside the parentheses as a squared expression:

y = 3((x + 2)^2 - 4) + 14

Finally, we can simplify this equation by distributing the 3 and combining like terms:

y = 3(x + 2)^2 - 2

So the vertex form of the quadratic equation y=3x^2 + 12x + 14 is y = 3(x + 2)^2 - 2. The vertex of this parabola is located at the point (-2, -2).

User Peter Tillemans
by
7.5k points
4 votes

Answer:

(2,-2)

Explanation:

User Roccer
by
8.9k points

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