Question

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

Answer 1

Answer:

1. Axis of symmetry: x = 3

2. Vertex: (3,5)

3. Solution of the equation:

• x-intercepts:

               $(3-\sqrt{5},0) \\\\(3+\sqrt{5},0)$

• y-intercept: (0, -4)

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:

       $x=\frac{-b\pm \sqrt{b^2-4ac} }{2a} \\\\x=\frac{-6\pm\sqrt{(-6)^2-4(-1)(-4)} }{2(-1)}\\\\x_1=3-\sqrt{5} \\\\x_2=3+\sqrt{5}$

5. Find the y-intercept

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

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