How do you graph the parabola ##y= (x + 3)^2 - 2## using vertex, intercepts and additional points?

Refer explanation section

The given quadratic equation is in the vertex form

##y=(x-3)^2-2##

Hence the vertex is ##(-3, -2)##

##(-3, -2)##This is on of the point on the curve.

##x=-3## is the minimum point on the curve. Hence to graph the curve, we take two point to the left of ##x=-3## and two point to its right.

Left side points -

At ##x=-4; y=(-4+3)^2-2=1-2=-1##

##(-4,-1)##

At ##x=-5; y=(-5+3)^2-2=4-2=2##

##(-5,2)##

Right side points.

At ##x=-2; y=(-2+3)^2-2=1-2=-1##

##(-2, -1)##

At ##x=-1; y=(-1+3)^2-2=4-2=2##

##(-1,2)##

Plot the points ##(-5, 2), (-4, -1), (-3, -2), (-2, -1), (-1, 2)##

You will get the graph.