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

User Fish
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1 Answer

4 votes

Answer:

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

User Jakes
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