Vertex Representation - Examples, Exercises and Solutions

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 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 $f(x) = 4x^2 + 4x$.

To convert the function $f(x) = (-x + 1)^2 + 3$ to its standard form, follow these steps:

Step 1: Expand the binomial $(-x + 1)^2$.
$(-x + 1)^2 = (-x)^2 + 2(-x)(1) + 1^2$

This simplifies to:
$(-x)^2 = x^2$
$2(-x)(1) = -2x$
$1^2 = 1$

Combining these terms gives:
$(-x + 1)^2 = x^2 - 2x + 1$

Step 2: Add the constant term $+3$ to the expanded form:
$f(x) = (x^2 - 2x + 1) + 3$

Step 3: Simplify the expression:
$f(x) = x^2 - 2x + 1 + 3 = x^2 - 2x + 4$

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

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

• Step 1: Expand $(x + 4)^2$ using the formula $(a + b)^2 = a^2 + 2ab + b^2$.
• Step 2: Simplify the expression by subtracting 16 from the expanded result.
• Step 3: Write the simplified expression in the standard form.

Now, let's work through each step:
Step 1: Start with the expression given in the problem:
$(x + 4)^2 = x^2 + 2 \cdot x \cdot 4 + 4^2$.

This results in:
$x^2 + 8x + 16$.

Step 2: Subtract 16 from the expanded expression:
$x^2 + 8x + 16 - 16 = x^2 + 8x$.

Step 3: The standard form of the expression is now:
$f(x) = x^2 + 8x$.

Therefore, the standard representation of the function is $f(x) = x^2 + 8x$.

To convert the function from vertex form to standard form, follow these steps:

• Step 1: Identify the vertex form - $f(x) = (x-2)^2 + 3$. The terms inside the parentheses represent a perfect square trinomial.
• Step 2: Expand the square. Recall: $(a-b)^2 = a^2 - 2ab + b^2$. Here, $a = x$ and $b = 2$.
• Step 3: Expand $(x-2)^2$:
  $(x-2)^2 = x^2 - 2 \cdot x \cdot 2 + 2^2 = x^2 - 4x + 4$.
• Step 4: Add the constant from the original function:
  $f(x) = (x^2 - 4x + 4) + 3 = x^2 - 4x + 7$.

After expanding and simplifying, we find that $f(x) = x^2 - 4x + 7$ is the standard form of the function.

Therefore, the correct choice that matches this solution is choice 3, which is $f(x) = x^2 - 4x + 7$.

To convert the quadratic function from vertex form to standard form, execute the following steps:

• Step 1: Begin with the given vertex form $f(x) = (x-5)^2 - 10$.
• Step 2: Expand $(x-5)^2$ using the formula $(a-b)^2 = a^2 - 2ab + b^2$, which results in:

$(x-5)^2 = x^2 - 2 \cdot x \cdot 5 + 5^2 = x^2 - 10x + 25$.

• Step 3: Replace the expanded form into the original function:

$f(x) = x^2 - 10x + 25 - 10$.

• Step 4: Combine like terms:

$f(x) = x^2 - 10x + 15$.

Therefore, the standard form of the function is $f(x) = x^2 - 10x + 15$.

Comparing with the given choices, the correct option is:

Choice 2: $f(x) = x^2 - 10x + 15$

To convert the given quadratic function into its standard form, follow these steps:

• Step 1: Expand the Binomial
  We begin with the function in vertex form: $f(x) = (x + 5)^2 + 3$. The expression $(x + 5)^2$ can be expanded using the binomial theorem: $(a + b)^2 = a^2 + 2ab + b^2$.

• Step 2: Apply the Expansion Formula
  Let $a = x$ and $b = 5$. Therefore, $(x + 5)^2 = x^2 + 2 \times x \times 5 + 5^2 = x^2 + 10x + 25$.

• Step 3: Add the Constant
  Now, add the constant 3 to this expanded result: $x^2 + 10x + 25 + 3 = x^2 + 10x + 28$.

Thus, the standard representation of the function is $f(x) = x^2 + 10x + 28$.

Given the choices, the correct answer is $f(x) = x^2 + 10x + 28$, which matches choice 2.

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 $(1, -1)$.

To solve for the vertex of the parabola given by the equation $y = (x-3)^2 - 1$, we start by comparing the equation with the standard vertex form of a quadratic function: $y = a(x-h)^2 + k$.

In the given equation, $y = (x-3)^2 - 1$, we identify:
- $h = 3$, which corresponds to the horizontal shift of the parabola.
- $k = -1$, which represents the vertical shift.

Therefore, the vertex of the parabola is at the point $(h, k)$, which is $(3, -1)$.

Thus, the vertex of the parabola is $(3, -1)$.

To solve for the vertex of the parabola given by the equation $y = x^2 + 3$, we will follow these steps:

• Step 1: Identify the coefficients from the equation. We have $a = 1$, $b = 0$, and $c = 3$.
• Step 2: Calculate the x-coordinate of the vertex using the formula $x = -\frac{b}{2a}$. Since $b = 0$, the formula becomes $x = -\frac{0}{2 \times 1} = 0$.
• Step 3: Find the y-coordinate of the vertex by substituting $x = 0$ back into the equation. Calculate $y = (0)^2 + 3 = 3$.

Therefore, the vertex of the parabola is at the point $(0, 3)$.

This corresponds to choice 3: $(0,3)$.

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 $(0, -6)$.

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 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 $(3, 0)$.

We need to convert the given function $f(x) = (x-2)^2 + 4$ to standard form.

To expand $(x-2)^2$, we use the formula $(a-b)^2 = a^2 - 2ab + b^2$. Applying this to $(x-2)^2$, we get:

• $(x-2)^2 = x^2 - 4x + 4$.

This accounts for the expanded square. Next, we add the constant term $4$ from the original function $(x-2)^2 + 4$:

• $f(x) = x^2 - 4x + 4 + 4$.

Simplify by combining the constant terms:

• $f(x) = x^2 - 4x + 8$.

The standard form of the function is thus $f(x) = x^2 - 4x + 8$.