Vertex Form Practice Problems - Quadratic Functions

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.

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

• Step 1: Identify the given information.
• Step 2: Apply the vertex formula to find $h$ and $k$.
• Step 3: Determine the coordinates of the vertex.

Now, let's work through each step:
Step 1: The given quadratic equation is $y = x^2 - 6$ where $a = 1$, $b = 0$, and $c = -6$.
Step 2: We use the vertex formula:

$h = -\frac{b}{2a}$
Substituting the values, $h = -\frac{0}{2 \cdot 1} = 0$.

$k = c - \frac{b^2}{4a}$
Using the given values, $k = -6 - \frac{0^2}{4 \cdot 1} = -6$.

Step 3: Therefore, the vertex of the parabola is at $(h, k) = (0, -6)$.

The solution to the problem is $(0, -6)$.

To solve this problem, let's identify the vertex of the given parabola in the form $y = (x - 3)^2$.

The parabola is already given in the vertex form of $y = (x - h)^2 + k$, which is a special case of the quadratic equation where the vertex ($h, k$) can be read directly from the equation.

• Step 1: We compare the given equation $y = (x - 3)^2$ with the standard vertex form $y = (x - h)^2 + k$.
• Step 2: Notice that $h = 3$ and since there is no additional term added or subtracted outside the squared term, $k = 0$.

Therefore, the vertex of the quadratic function $y = (x - 3)^2$ is $(3, 0)$.

The correct choice from the multiple-choice options provided is the one that matches $(3, 0)$.

The solution to the problem is the vertex $(3, 0)$.

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 $(1, -1)$.