The general "vertex form" for a parabola (in standard position) is
color(white)("XXX")y=color(green)m(x-color(red)a^2)+color(blue)bXXXy=m(x−a2)+b
with vertex at (color(red)a,color(blue)b)(a,b)
Notice that the given equation
color(white)("XXX")y=(x+1)^2-4XXXy=(x+1)2−4
is almost in this form, and we could rewrite it as
color(white)("XXX")y=color(green)1(x-color(red)(""(-1)))^2+color(blue)(""(-4))XXXy=1(x−(−1))2+(−4)
with vertex at (color(red)(-1,color(blue)(-4)))(−1,−4)
The yy-intercept is the value of yy when x=0x=0
and using the given equation:
color(white)("XXX")y_(x=0) =(0+1)^2-4 =-3XXXyx=0=(0+1)2−4=−3
So (0,-3)(0,−3) is a point on the parabola.
Note that the axis of symmetry (for a parabola in standard position) is a vertical line (i.e. x=x= some constant) through the vertex;
so in this case the axis of symmetry is x=color(red)(-1)x=−1.
If the axis of symmetry is x=-1x=−1 and (0,-3)(0,−3) is a point on the parabola,
since (0,-3)(0,−3) is a point 11 unit to the right of the vertical line x=-1x=−1
then there is another point 11 unit to the left of x=-1x=−1 with the same yy coordinate, namely (-2,-3)(−2,−3)
The three points (-1,-4), (0,-3), and (-2,-3)(−1,−4),(0,−3),and(−2,−3) should be sufficient to sketch the parabola (although, if you like, you could solve the given equation for the x-intercept values as well by setting y=0# in the original equation):
![enter image source here]()
