Answer:
The vertex is (3,-16) and axis of symmetry is 3
Step-by-step explanation:
The vertex of a quadratic equation \(f(x)=ax^{2}+bx+c\) is calculated as \((\frac{-b}{2a},f(\frac{-b}{2a}))\)
and the axis of symmetry is \(\frac{-b}{2a}\)
Compare equation \(y=x^{2}-6x-7\) with \(f(x)=ax^{2}+bx+c\)
Where a= 1 and b = -6
so, \(x=\frac{-b}{2a}=\frac{-(-6)}{2}\)
\(x=3\)
Now, put the value of x=3 in \(y=x^{2}-6x-7\)
\(y=(3)^{2}-6(3)-7\)
\(y=9-18-7\)
\(y=-16\)
Therefore, the vertex is (3,-16) and axis of symmetry is 3
