algebra question?

For a quadratic function: f(x) = a(x - h)² + k²
If a > 0, the graph of the quadratic function opens upward.
If a < 0, the graph of the quadratic function opens downward.

f(x) = 2[x² + 3x] - 7
f(x) = 2[x² + 2(3/2)x + (3/2)²] - 2(3/2)² - 7
f(x) = 2[x + (3/2)]² - (23/2)
Since 2 > 0, the graph opens upward.

For y = ax² + bx + c, its upwards iff a > 0. 

So in your case, is a > 0 or < 0 ?
Conclusion?

Done!

Don't forget to vote me best answer for being the first to correctly walk you through without spoiling out the answer. That way it gives you a chance to workout it and to get good at it.

 f(x) = 2x² + 6x – 7
 The graph opens upward parabola focus | (-3/2, -91/8) = (-1.5, -11.375) vertex | (-3/2, -23/2) = (-1.5, -11.5) semi-axis length | 1/8 = 0.125 focal parameter | 1/4 = 0.25 eccentricity | 1 directrix | f = -93/8