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

Jul 6, 2018

$x = 1 , \text{ vertex } = \left(1 , 1\right)$

#### Explanation:

$\text{find the zeros by letting } f \left(x\right) = 0$

$- {x}^{2} + 2 x = 0$

$x \left(2 - x\right) = 0$

$x = 0 \text{ and "x=2larrcolor(blue)"zeros}$

$\text{the x-coordinate of the vertex lies on the axis of symmetry}$
$\text{which is situated at the midpoint of the zeros}$

$\text{axis of symmetry is } x = \frac{0 + 2}{2} = 1$

$\text{substitute this value into the equation for y-coordinate}$

${y}_{\text{vertex}} = - 1 + 2 = 1$

$\textcolor{m a \ge n t a}{\text{vertex }} = \left(1 , 1\right)$
graph{-x^2+2x [-10, 10, -5, 5]}

Jul 6, 2018

Vertex is at $\left(1 , 1\right)$ , axis of symmetry is $x = 1$

#### Explanation:

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

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

$f \left(x\right) = - \left({x}^{2} - 2 x + 1\right) + 1$ or

$f \left(x\right) = - {\left(x - 1\right)}^{2} + 1$ Comparing with vertex form of

equation f(x) = a(x-h)^2+k ; (h,k) being vertex we find

here $h = 1 , k = 1 \therefore$ Vertex is at $\left(1 , 1\right)$

Axis of symmetry is x= h or x = 1 ;

graph{-x^2+2 x [-10, 10, -5, 5]} [Ans]