SOLUTION
[tex]\begin{gathered} g(x)=-5(x-3)^2 \\ \text{when g(x)=0} \\ 0\text{ }=-5(x-3)^2 \\ 0=-5(x-3)(x-3) \\ \text{Hence , }x=3\text{ (twice)} \\ Therefore,x\text{ intercept = (3,0)} \end{gathered}[/tex]
2.
[tex]\begin{gathered} g(x)=-5(x-3)^2+0 \\ \text{EQUATION OF A VERTEX} \\ Y=a(x-h)^2+k \\ \text{comparing the equations} \\ h=3\text{ and k=0} \\ \text{Therefore, the vertex is (3,0)} \end{gathered}[/tex]
