Question

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

Answer 1

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

Answer 2

i still dont understand

Answer 3

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

Answer 4

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 \(-\frac{b}{2a}\)

Answer 5

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

Answer 6

so x of the vertex is 2?