Question

Y=-x^2+6x-4
Find the Axis of Symmetry and the Vertex, Also solve the whole equation.

Answer 1

1. Axis of symmetry: x = 3

2. Vertex: (3,5)

3. Solution of the equation:

• x-intercepts:

`(3-√(5),0) \\\\(3+√(5),0)`

• y-intercept: (0, -4)

Step-by-step explanation:

1. Equation:

`y=-x^2+6x-4`

2. Axis of symmetry:

That is the equation of a parabola, whose standard form is:

`y=ax^2+bx+c`

Where:

`a=-1;b=6;c=-4`

The axis of symmetry is the vertical line with equation:

`x=-b/2a`

Substitute
`a=-1,\text{ and }b=6`

`x=-6/[(2)(-1)]=-6/(-2)=3`

Thus, the axis of symmetry is:

`x=3`

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:

`y=-x^2+6x-4\\\\y=-(3)^2+6(3)-4=-9+18-4=5`

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.

`y=-x^2+6x-4=0`

Use the quadratic equation:

`x=(-b\pm √(b^2-4ac) )/(2a) \\\\x=(-6\pm√((-6)^2-4(-1)(-4)) )/(2(-1))\\\\x_1=3-√(5) \\\\x_2=3+√(5)`

5. Find the y-intercept

The y-intercet is the value of y when x=0:

`y=-x^2+6x-4\\\\y=-(0)^2+6(0)-4=-4`