Find the Axis of Symmetry in Quadratic Equation -5x² + 10

Given the expression of the quadratic function

Finding the symmetry point of the function

$f(x)=-5x^2+10$

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 $(0,10)$.