# How do you find the axis of symmetry and vertex point of the function: f(x) = -x^2 + 10?

Oct 7, 2015

The axis of symmetry is $x = 0$.
The vertex is $\left(0 , 10\right)$.

#### Explanation:

$f \left(x\right) = - {x}^{2} + 10$

Substitute $y$ for $f \left(x\right)$.

$y = - {x}^{2} + 10$ is a quadratic equation in the form $a {x}^{2} + b x + c$, where $a = - 1 , b = 0 , c = 10$.

Axis of symmetry
The axis of symmetry is an imaginary horizontal line that divides the parabola into two equal halves.

The formula for finding the axis of symmetry is $x = \frac{- b}{2 a}$

$x = \frac{- b}{2 a} = \frac{0}{2 \cdot - 1} = 0$

The axis of symmetry is $x = 0$.

Vertex
The vertex is the maximum or minimum point on the parabola. In this case, because $a$ is negative, the parabola opens downward and so the vertex is the maximum point.

The $x$ value of the vertex is $0$.

To find the $y$ value of the vertex is determined by substituting the $x$ value into the equation and solving for $y$.

$y = - {\left(0\right)}^{2} + 10 = 10$

The vertex is $\left(0 , 10\right)$.

graph{y=-x^2+10 [-16.82, 15.2, -4.1, 11.92]}