Answer:
1. Axis of symmetry: x = 3
2. Vertex: (3,5)
3. Solution of the equation:
[tex](3-\sqrt{5},0) \\\\(3+\sqrt{5},0)[/tex]
Explanation:
1. Equation:
[tex]y=-x^2+6x-4[/tex]
2. Axis of symmetry:
That is the equation of a parabola, whose standard form is:
[tex]y=ax^2+bx+c[/tex]
Where:
[tex]a=-1;b=6;c=-4[/tex]
The axis of symmetry is the vertical line with equation:
[tex]x=-b/2a[/tex]
Substitute [tex]a=-1,\text{ and }b=6[/tex]
[tex]x=-6/[(2)(-1)]=-6/(-2)=3[/tex]
Thus, the axis of symmetry is:
[tex]x=3[/tex]
3. Vertex
The x-coordinate of the vertex is equal to the axys of symmetry, i.e x = 3.
To find the y-xoordinate, substitute this value of x into the equation for y:
[tex]y=-x^2+6x-4\\\\y=-(3)^2+6(3)-4=-9+18-4=5[/tex]
Therefore, the vertex is (3, 5)
4. Find the x-intercepts
The x-intercepts are the roots of the equation, which are the points wher y = 0.
[tex]y=-x^2+6x-4=0[/tex]
Use the quadratic equation:
[tex]x=\frac{-b\pm \sqrt{b^2-4ac} }{2a} \\\\x=\frac{-6\pm\sqrt{(-6)^2-4(-1)(-4)} }{2(-1)}\\\\x_1=3-\sqrt{5} \\\\x_2=3+\sqrt{5}[/tex]
5. Find the y-intercept
The y-intercet is the value of y when x=0:
[tex]y=-x^2+6x-4\\\\y=-(0)^2+6(0)-4=-4[/tex]
