Given:
[tex]y=-3x^2+12x-9[/tex]Required:
We need to find the vertex and the point lies on the given parabola.
Explanation:
The given equation is of the form.
[tex]y=ax^2+bx+c[/tex]where a =-3, b =12 and c =-9.
Consider the formula to find the x-coordinate of the vertex.
[tex]x=-\frac{b}{2a}[/tex]Substitute a =-3 and b =12 in the formula.
[tex]x=-\frac{12}{2(-3)}[/tex][tex]x=-\frac{12}{-6}[/tex][tex]x=2[/tex]Substitute x =2 in the given equation to find the y-coordinate of the vertex.
[tex]y=-3(2)^2+12(2)-9[/tex][tex]y=-12+24-9[/tex][tex]y=3[/tex]The vertex is (2,3).
Substitute x =0 in the given equation.
[tex]y=-3(0)^2+12(0)-9[/tex][tex]y=-9[/tex]The point is (0,-9).
Substitute x =1 in the given equation.
[tex]y=-3(1)^2+12(1)-9[/tex][tex]y=-3+12-9[/tex][tex]y=0[/tex]The point is (1,0).
Final answer:
Vertec at (2,3) and passes through (1,0).