vertex has x coordinate = 1
It's an upward opening parabola with general quadratic equation, in vertex form
: y = a(x-h)^2 + k where h=1 = the x coordinate of the vertex and k=-4 = the y coordinate of the vertex. a>0
y = a(x-1)^2 -4
x=-1 is an x intercept and since x=1 is the axis of symmetry there's another x intercept an equal distance from x=1 on the other side, i.e. x=3 is the other x intercept, each x intercept a distance of 2 from the axis of symmetry
y = a(x+1)(x-3) is another form of the parabola's equation
y =a(x^2 -2x -3) along with y =a(x-1)^2 -4= a(x^2 -2x + 1) -4
set them equal and solve for a
ax^2 -2ax -3a = ax^2 -2ax +1 -4
-3a =-3
a = 1
the parabola's equation is
y =(x+1)(x-3) in x intercept form
or
y = x^2 -2x -3 in standard quadratic form:
y=ax^2 +bx + c with a=1, b=-2, and c=-3= y intercept
or
y = (x-1)^2 -4 in vertex form
the y intercept = -3 or the point (0, -3) It's the "c" in the y=ax^2 + bx + c form of the equation.
it might help to plot the points, the vertex, the x intercepts and a rough sketch of the parabola
