Question

1. which graph represents the equation $-4x^{2}+8y + 7 = 0$?

Explanation:

Step1: Rearrange to vertex form

Start with the given equation:
$-4x^2 + 8y + 7 = 0$
Isolate the $y$-term:
$8y = 4x^2 - 7$
Divide by 8:
$y = \frac{4}{8}x^2 - \frac{7}{8}$
Simplify:
$y = \frac{1}{2}x^2 - \frac{7}{8}$

Step2: Identify conic type/shape

The equation $y = \frac{1}{2}x^2 - \frac{7}{8}$ is a quadratic in $x$, so it is a parabola. Since the coefficient of $x^2$ is positive, it opens upward. The vertex is at $(0, -\frac{7}{8})$, which is slightly below the origin.

• The first graph is a hyperbola (not a parabola), so it is incorrect.
• The second graph is a downward-opening parabola, which does not match the positive coefficient of $x^2$.

(Note: Assuming the full question includes an upward-opening parabola option that is the correct match, as the provided options are incomplete. Based on the derived equation, the correct graph is an upward-opening parabola with vertex at $(0, -\frac{7}{8})$.)

Answer:

The correct graph is an upward-opening parabola with vertex at $(0, -\frac{7}{8})$ (matching the equation $y = \frac{1}{2}x^2 - \frac{7}{8}$; the provided partial options do not include this, so the intended correct option is the upward-opening parabola not fully shown here).