Vertex form of the quadratic equation

To solve this problem, we'll perform these steps:

• Expand $(x-3)^2$ using the formula for a square:
  $(x-3)^2 = x^2 - 2 \times x \times 3 + 3^2 = x^2 - 6x + 9$
• Add the $x$ from $f(x) = (x-3)^2 + x$ to the expanded terms:
  $x^2 - 6x + 9 + x$
• Combine like terms:
  $x^2 - 6x + x + 9 = x^2 - 5x + 9$

Therefore, the standard form of the function $f(x)$ is $f(x) = x^2 - 5x + 9$.

Thus, the correct choice is Choice 3.

The equation $y = (x+1)^2$ is already in the vertex form $y = (x-h)^2 + k$, where $(h, k)$ is the vertex of the parabola.

By comparing, we have:

• The expression inside the square is $(x+1)$, which can be rewritten as $(x - (-1))$. Thus, $h = -1$.
• The term $k$ is not present, which means $k = 0$.

Therefore, the vertex $(h, k)$ of the parabola is $(-1, 0)$.

Thus, the correct answer is $(-1, 0)$.

The given equation is $y = (x-1)^2 - 1$. This equation is in the vertex form, $y = a(x-h)^2 + k$, where $a$, $h$, and $k$ are constants.

In this case, the given equation can be written as $y = 1 \cdot (x-1)^2 + (-1)$, indicating that $a = 1$, $h = 1$, and $k = -1$.

The vertex form of the quadratic equation allows us to directly identify the vertex of the parabola as $(h, k)$.

From our identification, it is clear that the vertex $(h, k)$ of the parabola is $(1, -1)$.

Therefore, the vertex of the given parabola is $(1, -1)$.

The given equation of the parabola is $y = (x+1)^2 - 1$.

This equation is already in the vertex form, $y = a(x-h)^2 + k$, where $(h, k)$ is the vertex.

By comparing, we identify:
The expression $(x + 1)$ implies that $h = -1$ (since $x + 1$ is equivalent to $(x - (-1))$).
The constant $-1$ is the $k$ value.

Thus, the vertex $(h, k)$ is $(-1, -1)$.

Therefore, the vertex of the parabola is at the point $(-1,-1)$.

To convert $f(x) = (2x+1)^2 - 1$ into its standard quadratic form, we need to expand $(2x+1)^2$ first and then adjust for the subtraction of 1.

The expansion is carried out using the binomial expansion formula:

$(2x + 1)^2 = (2x)^2 + 2(2x)(1) + 1^2$.

Calculating each term gives:

• $(2x)^2 = 4x^2$
• $2(2x)(1) = 4x$
• $1^2 = 1$

Combining these, we obtain:

$(2x + 1)^2 = 4x^2 + 4x + 1$

Now, substituting back into the original equation:

$f(x) = (2x+1)^2 - 1 = (4x^2 + 4x + 1) - 1$

Subtracting 1 from the constant term, we get:

$f(x) = 4x^2 + 4x + 1 - 1 = 4x^2 + 4x$

Therefore, the standard form representation of the function is $f(x) = 4x^2 + 4x$.