# What is the vertex of y=-x^2+6?

Mar 6, 2017

$\left(0 , 6\right)$

#### Explanation:

Consider the standardised form of $y = a {x}^{2} + b x + c$

Written as $y = a \left({x}^{2} + \frac{b}{a} x\right) + c$

x_("vertex")=(-1/2)xxb/a" "->" "(-1/2)xx0/(-1) = 0

The y-intercept$= c = 6$

As there is no $b x$ term in $y = - {x}^{2} + 6 \text{ }$ the axis of symmetry is the y-axis. So the vertex is at $\left(x , y\right) = \left(0 , 6\right)$

As the ${x}^{2}$ term is negative then the general shape of the curve is $\cap$