Graphing Parabolas using the vertex?

Question

A. y = -2(x + 1)^2 - 3

B. y= 3(x + 4)^2 - 6

Identify the vertex of each. Then sketch the graph.

1. Plot the point

2. Find a second point

3. Find the third point using symmetry

Answer 1

y=-2(x+1)^2 -3 has vertex (-1,3) the general equation is y=a(x-h)^2+k where (h,k) is the vertex. that point is in quadrant 2, start at the origin and go left 1 and up 3. the vertex is the maximum point of the parabola. It's downward opening

Let x=0, then y=-2-3 = -5, the point (0,-5) is on the parabola. It has axis of symmetry x=-1, so another point on the parabola is (-2,-5). (0,5) is one unit to the right of the axis of symmetry. (-2,-5) is one unit to the left of the axis of symmetry

y=3(x+4)^2 - 6 has vertex and minimum point (-4,-6) in quadrant III.

It's an upward opening parabola

Axis of symmetry is x=-4

let x=0, then y =48-6 = 42 (0,42) is a point on the parabola, so is (-8,42)