Y = x^2 -3
y = (x-3)(x+3)
For x-intercepts, y = 0
Therefore (x-3)(x+3) = 0
x intercepts are (3,0) and (-3,0)
The line of symmetry is x = 3+(-3) = 0
Vertex is on the line of symmetry
When x = 0, y = -1
Vertex = (0,-1)
Answer:2x
Step-by-step explanation:
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Answer:
- (x -3)^2 +(y -5)^2 = 9
- x^2 +y^2 -6x -10y +25 = 0
Step-by-step explanation:
The standard form equation for a circle of radius r and center (h, k) is ...
(x -h)^2 +(y -k)^2 = r^2
For radius 3 and center (3, 5), this is ...
(x -3)^2 +(y -5)^2 = 9 . . . . standard form equation
Expanding this and subtracting the constant on the right gives ...
x^2 -6x +9 +y^2 -10y +25 -9 = 0
x^2 +y^2 -6x -10y +25 = 0 . . . . general form equation