err... complete the square???

Question

help me please, i know this is not hard but my head is not running = ="

complete the square to find the vertex and axis of symmetry

f(x) = 2x^2 + 12x +15

Answer 1

f(x) = 2x^2 + 12x + 15.

When completing the square, your first step is to factor out the coefficient of x^2 out of the first two terms.

f(x) = 2(x^2 + 6x) + 15

Now, this is the tricky part. What you want to do is add a number within the brackets to make the polynomial a perfect square. To obtain this magic number, you have to take "half squared" of the coefficient of x.

When I say "half squared", I mean, you multiply 6 by 1/2, take the result, and square it. Half of 6 is 3, squared is 9. So you would add 9 within the brackets.

f(x) = 2(x^2 + 6x) + 15
f(x) = 2(x^2 + 6x + 9) + 15 + ?

Note that I put a question mark there. The reason why I did it was because we can't just add a number in an equation out of nowhere; we have to compensate for it.

Notice that we added the number 9 within the brackets. Does that mean we can subtract 9 to offset the 9 we added? The answer is no, because what we REALLY did was add 2(9), or 18. When we added the 9 WITHIN the brackets, that lingering 2 that is being multiplied to the brackets means we actually added 18 to the equation. To compensate for that, we have to SUBTRACT 18.

f(x) = 2(x^2 + 6x + 9) + 15 + ?
f(x) = 2(x^2 + 6x + 9) + 15 - 18

Now, our bracketed terms contains a perfect square. Also, we can merge 15 - 18 into -3.

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

This is now in the vertex form,

f(x) = A(x - h)^2 + k, where (h,k) is the coordinates of the vertex.

In our case, to find the vertex we take the NEGATIVE of the number inside of the brackets (it's 3, so our x-coordinate is -3), and then we take the number k normally (which is -3).
Therefore the vertex is at (-3, -3).

Answer 2

We have the equation

2x^2 + 12x + 15

To complete the square, we want to factor the first two terms. We can factor out the 2:

2(x^2 +6x ____) +15

To complete the square, we want to look at the coeffient of the 'x' variable, which is '6'. We first want to take half of 6, which is 3, and then square it, which would be 9. After doing so, we are left with the equation:

2(x^2 + 6x + 9) + 15

We want to make sure we do not change the value of the equation, so we have to do the same to the other term. SInce we put in a '9' when we factored, we must change the term '15' so we do not change the value. In essence we are adding a '18', so we must SUBTRACT an '18' from 15.

2(x^2 +6x + 9) + 15 - 18

We then factor our polynomial and add our 15 and -18 leaving us with:

2(x + 3)^2 -3

To find the vertex of the parabola, we will use the function:

a(x-h)^2 + k

The vertex will represent (h,k)

The vertex is (-3, -3).

To find the axis of symmetry, we will use the function f(-x) where we will substitute an (-x) for (x)

2(-x)^2 + 12(-x) + 15

This will simplify giving us:
2x^2 - 12x + 15

Since there is no change of all signs from the original equation to the equation we got from f(-x), there is no symmetry.

Hope that helps.

Answer 3

Okay, first take out the leading coefficient from the first two terms.

f(x) = 2 (x^2 + 6x + ____) + 15

Now see how to make the expression in () a perfect square. And be sure to take away whatever you added to the function.

f(x) = 2 (x^2 + 6x + 9) + 15 - 18

Simplify.

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

The form is now in f(x) = a(x + h)^2 - k.

The vertex is (h,k) or (3 , 3).
The axis of symmetry is x = -3.

Answer 4

This is an equation of a parabola .
Vertex should be derived by differentiation with respect to x,
that is = 4 x +12
and equalising the result to zero.
gives x (of the vertex ) = -3,
then y = -3
Vertex (-3, -3)
axis of symetry is vertical, I mean parallel to the Y axis which is
x = -3

Answer 5

Are you sure you copied the problem correctly? Please check your book. In order to complete the square, you need to have this type of format: Ax^2 + By^2 + Ax + By - C = 0

Answer 6

ahh!! MATH!!! RUNAWAYYY