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Convert The Equation Y=X^(2)+4x+11 Into Vertex Form

User Jamian
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The equation we are given is in standard form, y = x^2 + 4x + 11. Our goal is to convert it into vertex form. The general form of vertex format is y = a(x - h)^2 + k, where (h,k) are the coordinates of the vertex of the parabola.

Step 1:
In our equation, 'a' is 1 (since there is no other number being multiplied to x^2). The task requires finding the values of 'h' and 'k'.

Step 2:
To find 'h' we take the coefficient of x, which is 4 in our case, divide it by 2 and take the opposite sign. So, h = -4/2 = -2.

Step 3:
Our equation now looks like this: y = (x + 2)^2 + k.

Step 4:
Now we need to find the value of 'k' to make sure that our equation remains the same. To find 'k', we can open the brackets from the above equation, which results in y = x^2 + 4x + 4 + k.

Step 5:
As this should be equal to our original equation, k must be the value that makes the constant term from this opened equation equal to the constant term in our original equation. Hence, 4 + k = 11, which gives k = 11 - 4 = 7.

So, the equation in vertex form is given by: y = (x + 2)^2 + 7.

This vertex form of the equation tells us that the parabola has its vertex at (-2, 7) and opens upwards (which is defined by a = 1 > 0).

Therefore, the vertex form of the equation Y=X^(2)+4x+11 is (-2, 7)

User Balualways
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