The equation y - 2 = 3(x + 1) is in point-slope
2 answers:
You might be interested in
x-intercepts: (-3,0), (1,0)
work:
0 = x^2 + 2x - 3
quadratic equation: x = -2 +√(2^(2) - 4 × 1(-3)) / (2 × 1) = 1
quadratic equation: x = -2 -√(2^(2) - 4 × 1(-3)) / (2 × 1) = -3
(x,y) --) (-3,0), (1,0)
y-intercept: (0,-3)
work:
y = (0)^2 + 2(0) - 3
y = -3
(x,y) --) (0,-3)
vertex: (-1,-4)
work:
xv = -( b / (2a) )
a = 1, b = 2, c = -3
xv = -( 2 / (2×1) )
xv = -1
yv = (-1)^2 + 2(-1) - 3
yv = -4
(x,y) --) (-1,-4)
axis of symmetry: -1
work:
a = 1 in the x^(2) + 2ax + a^(2)
x^(2) + 2x + 1^(2) = (x + 1)^(2)
(x + 1)^(2) - 3 - 1^(2)
y = (x + 1)^(2) - 4
y + 4 = (x - 1)^(2)
put in standard form --) 4 × 1/4( y -( -4 ) ) = ( x -( -1) )^(2)
(h,k) = (-1,-4); p = 1/4
in parabola form expression: 4p( y - k ) = ( x - h )^(2) and is symmetric around the y-axis at -1.
