find the zeros and the vertex of the equation y=x^2-4x-5

Question

anonymous · Answer

first coordinate of the vertex is alway $-\frac{b}{2a}$ which in your case is $-\frac{-4}{2}=2$
second coordinate of the vertex is what you get for $y$ when you replace $x$ by the first coordinate

anonymous · Answer

i still dont understand

anonymous · Answer

general quadratic looks like 
$$y=ax^2+bx+c$$ you have 
$$y=x^2-4x-5$$so in your example 
$$a=1,b=-4,c=-5$$

anonymous · Answer

the vertex is an ordered pair, looks like $(x, y)$ 
it is either the highest or lowest point on the parabola. the first coordinate of the vertex of $y=ax^2+bx+c$ is $-\frac{b}{2a}$

anonymous · Answer

in your example, $-\frac{b}{2a}=-\frac{-4}{2\times 1}=2$

anonymous · Answer

so x of the vertex is 2?