Answer:
1. Vertex: (3,4)
2. It opens down.
3. The y-intercept is -5
Step-by-step explanation:
1. For a quadratic function in the form [tex]y=ax^2+bx+c[/tex], you can calculate the x-coordinate of the vertex with:
[tex]x=\frac{-b}{2a}[/tex]
Then, given the function [tex]y=-x^2+6x-5[/tex], you can identify that:
[tex]b=6\\a=-1[/tex]
Substituting, you get:
[tex]x=\frac{-6}{2(-1)}=3[/tex]
Substitute [tex]x=3[/tex] into [tex]y=-x^2+6x-5[/tex] to find the y-coordinate of the vertex:
[tex]y=-(3)^2+6(3)-5=4[/tex]
Then the vertex is at (3,4)
2. You identified above the value of "a". This is:
[tex]a=-1[/tex]
Then, since [tex]a<0[/tex], the parabola opens down.
3. Substitute [tex]x=0[/tex] into the function and solve for "y":
[tex]y=-(0)^2+6(0)-5\\y=-5[/tex]
Therefore, the y-intercept is -5
