Symmetry: Finding a stationary point

To determine the symmetry (vertex) point of the quadratic function $f(x) = \frac{1}{2}x^2$, we will use the formula for the x-coordinate of the vertex (or axis of symmetry) for a general quadratic function $f(x) = ax^2 + bx + c$, which is given by:

• $x = -\frac{b}{2a}$

In this problem, the coefficients are $a = \frac{1}{2}$, $b = 0$, and $c = 0$. By substituting these values into the vertex formula:

$x = -\frac{0}{2 \times \frac{1}{2}} = 0$

This tells us that the x-coordinate of the vertex is $0$. To find the y-coordinate of the vertex, we substitute $x = 0$ back into the function $f(x) = \frac{1}{2}(x^2)$:

$f(0) = \frac{1}{2}(0)^2 = 0$

Thus, the vertex of the function, also its symmetry point, is at the coordinate $(0,0)$.

Therefore, the symmetry point of the function $f(x) = \frac{1}{2}x^2$ is $(0, 0)$.

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

• Identify the coefficients of the quadratic function.
• Apply the vertex formula to find the axis of symmetry and subsequently the vertex.
• Calculate the function's value at the symmetry point.

Now, let's work through each step:
Step 1: The given function is $f(x) = 2x^2$, where $a = 2$ and $b = 0$.
Step 2: The axis of symmetry for a quadratic function in the form $ax^2 + bx + c$ is given by $x = -\frac{b}{2a}$. With $b = 0$, this simplifies to $x = 0$.
Step 3: To find the vertex, calculate the function's value at $x = 0$, using $f(x) = 2x^2$.
Plugging in $x = 0$, we find:
$f(0) = 2(0)^2 = 0$.
Thus, the vertex, and hence the symmetry point of the function, is $(0, 0)$.

Therefore, the solution to the problem is $(0, 0)$.

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

• Step 1: Identify the coefficients $a$, $b$, and $c$ from the quadratic function.
• Step 2: Apply the vertex formula to find the x-coordinate of the symmetry point.
• Step 3: Substitute the x-coordinate back into $f(x)$ to find the y-coordinate of the vertex.
• Step 4: Conclude the symmetry point as the vertex, $(x, f(x))$.

Now, let's work through each step:

Step 1: The quadratic function is $f(x) = -5x^2 + 10$. The coefficients are $a = -5$, $b = 0$, and $c = 10$.

Step 2: Applying the vertex formula $x = -\frac{b}{2a}$, we have:

$x = -\frac{0}{2(-5)} = 0$.

Step 3: Substitute $x = 0$ back into the function:

$f(0) = -5(0)^2 + 10 = 10$.

Step 4: Therefore, the vertex and symmetry point of the function is $(0, 10)$.

The correct choice from the given options is $(0,10)$.

Therefore, the solution to the problem is $(0,10)$.

To find the symmetry point of the quadratic function $f(x) = 3 - 5x^2$, we follow these steps:

• Identify that the function is in the form $f(x) = ax^2 + bx + c$, where $a = -5$, $b = 0$, and $c = 3$.

• The x-coordinate of the symmetry point, also known as the vertex, is given by the formula $x = -\frac{b}{2a}$.

• Substitute the values: $x = -\frac{0}{2(-5)} = 0$.

• Calculate the y-coordinate by substituting $x = 0$ into the function: $f(0) = 3 - 5(0)^2 = 3$.

• Hence, the symmetry point of the function is $(0, 3)$.

Therefore, the symmetry point of the function $f(x) = 3 - 5x^2$ is $(0, 3)$.

To find the symmetry point of the quadratic function $f(x) = 3x^2 + 6x$, follow these steps:

• Step 1: Identify the coefficients from the quadratic function $ax^2 + bx + c$. Here, $a = 3$ and $b = 6$.
• Step 2: Use the vertex formula for the symmetry point, which is given by $x = -\frac{b}{2a}$.
• Step 3: Substitute the values of $a$ and $b$ into the formula:

$x = -\frac{6}{2 \times 3} = -\frac{6}{6} = -1$

• Step 4: Substitute $x = -1$ back into the original function to find the corresponding $y$-coordinate:

$f(-1) = 3(-1)^2 + 6(-1) = 3 \times 1 - 6 = 3 - 6 = -3$

• Step 5: Therefore, the symmetry point of the function is $(-1, -3)$.

Thus, the symmetry point of the given quadratic function $f(x) = 3x^2 + 6x$ is $\boxed{(-1, -3)}$.

To solve this problem, we need to find the symmetry point of the quadratic function given by $f(x) = 3 + 3x^2$.

• Step 1: Identify the type of quadratic equation. Here, it's $ax^2 + bx + c$ where $a = 3$, $b = 0$, and $c = 3$.
• Step 2: Use the formula for the symmetry point $x = -\frac{b}{2a}$. Since $b = 0$, the formula simplifies to $x = 0$.
• Step 3: Calculate the y-coordinate by substituting $x = 0$ into the original function: $f(0) = 3(0)^2 + 3 = 3$.

Thus, the symmetry point (also the vertex of the parabola) is $(0, 3)$.

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

To find the symmetry point of the quadratic function $f(x) = -4x^2 + 8x + 3$, follow these steps:

• Step 1: Identify key parameters
  The function is of the form $f(x) = ax^2 + bx + c$, with $a = -4$, $b = 8$, and $c = 3$.
• Step 2: Find the x-coordinate of the vertex
  Use the formula for the x-coordinate of the vertex: $x = -\frac{b}{2a}$.
  Substitute the values for $b$ and $a$:
  $x = -\frac{8}{2 \times -4} = -\frac{8}{-8} = 1$.
• Step 3: Find the y-coordinate by substituting back into $f(x)$
  Calculate $f(1)$:
  $f(1) = -4(1)^2 + 8(1) + 3 = -4 + 8 + 3 = 7$.
• Step 4: State the symmetry point
  The symmetry point, or vertex, of the function is $(1, 7)$.

Therefore, the symmetry point of the quadratic function $f(x) = -4x^2 + 8x + 3$ is $(1, 7)$.

To find the symmetry point of the quadratic function $f(x) = 5x - x^2$, we will determine the vertex of the parabola.

• Step 1: Express the quadratic function in standard form:
  The given function is already in standard form: $f(x) = -x^2 + 5x$, where $a = -1$ and $b = 5$.

• Step 2: Apply the vertex formula to find the x-coordinate of the vertex:
  For the quadratic function $ax^2 + bx + c$, the x-coordinate of the vertex is found using $x = -\frac{b}{2a}$.

• Step 3: Calculate the x-coordinate:
  $\begin{aligned} x &= -\frac{5}{2 \times (-1)} \\ &= -\frac{5}{-2} \\ &= \frac{5}{2} \end{aligned}$

• Step 4: Substitute $x = \frac{5}{2}$ back into the function to find the y-coordinate:
  $\begin{aligned} f\left(\frac{5}{2}\right) &= -\left(\frac{5}{2}\right)^2 + 5\left(\frac{5}{2}\right) \\ &= -\frac{25}{4} + \frac{25}{2} \\ &= -\frac{25}{4} + \frac{50}{4} \\ &= \frac{25}{4} \end{aligned}$

• Step 5: Determine the symmetry point:
  The symmetry point, and thus the vertex of the function, is $\left(\frac{5}{2}, \frac{25}{4}\right)$, or $(2\frac{1}{2}, 6\frac{1}{4})$.

Therefore, the symmetry point of the function is $(2\frac{1}{2}, 6\frac{1}{4})$.

To solve this problem, we'll find the vertex of the quadratic function $f(x) = -6x^2 + 24x$.

• Step 1: Identify the coefficients from the quadratic function.
• Step 2: Use the vertex formula to find the x-coordinate.
• Step 3: Substitute the x-coordinate back into the function to find the y-coordinate.

Now, let's work through each step:
Step 1: The function given is $f(x) = -6x^2 + 24x$. Here, $a = -6$ and $b = 24$.
Step 2: Use the vertex formula $x = -\frac{b}{2a}$. Substituting the values, we get $x = -\frac{24}{2 \cdot (-6)} = \frac{24}{12} = 2$.
Step 3: Substitute $x = 2$ back into the function to find the y-coordinate: $f(2) = -6(2)^2 + 24 \cdot 2 = -24 + 48 = 24$.

Therefore, the symmetry point of the function is $(2, 24)$.

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

• Step 1: Identify the given coefficients from $f(x) = ax^2 + bx + c$: $a = -3$, $b = 0$, and $c = 12$.
• Step 2: Apply the vertex formula $x = -\frac{b}{2a}$ to find the x-coordinate of the vertex.
• Step 3: Plug the x-coordinate back into the function to find the y-coordinate.

Now, let's work through each step:
Step 1: The function is given as $f(x) = -3x^2 + 12$. We have $a = -3$, $b = 0$, and $c = 12$.
Step 2: Using the formula $x = -\frac{b}{2a}$, substitute $b = 0$ and $a = -3$:
$x = -\frac{0}{2 \times -3} = 0$
Step 3: Substitute $x = 0$ back into the quadratic function to find the y-coordinate:
$f(0) = -3(0)^2 + 12 = 12$
So, the vertex, or symmetry point, of the function is $(0, 12)$.

Therefore, the solution to the problem is $(0, 12)$.