Please help if you know how to do this, thanks

9514 1404 393

  1.  see the first attachment for a graph; quadrants I, II, III; vertex (-3, -4); x-intercepts -5, -1; graph is dashed line, shaded inside the parabola, which opens up

  2.  see the second attachment for a graph; all quadrants; vertex (1, 2); x-intercepts 1±√2; graph is a solid line, shaded outside the parabola, which opens down; (-7, 0) is a solution

Step-by-step explanation:

1. The boundary expression is in vertex form, so the vertex can be read directly.

  y = (x -h)^2 +k . . . . . . parabola with vertex (h, k)

  y > (x +3)^2 -4 . . . . .  (h, k) = (-3, -4)

The leading coefficient is +1, a positive number, so the parabola opens up. The relation is y > ( ), so values of y above the dashed line will be shaded. Those values are inside the parabola. The points (0, 0) and (-7, 0) are not in the shaded area, so are not solutions.

__

2. Again, the boundary expression is in vertex form, so the vertex can be read directly.

  y ≥ -(x -1)^2 +2 . . . . . (h, k) = (1, 2)

The x-intercepts are found by solving for x when y=0:

  0 = -(x -1)^2 +2

  x -1 = ±√2

  x = 1±√2

The leading coefficient is -1, so the parabola opens downward. The relation is y ≥ ( ), so y-values above the solid line will be shaded. Those values are outside the parabola. Point (0, 0) is not a solution. Point (-7, 0) is a solution.

_____

Additional comment

As with the graph of any inequality, the boundary line is solid when the "or equal to" case is part of the relation. If the relation is strictly "greater than" or "less than", then the boundary line is dashed.