Match the equations to the graphs. A) y=x^2+8x+12, Б) y=x^2-8x+12, B) y=-x^2+8x-12

Matching Graphs to Equations

We need to match the given quadratic equations to their corresponding graphs. The key features to consider are the direction of the parabola (determined by the sign of the $$x^2$$ coefficient) and the location of the vertex.

Equation Analysis:

• A) $$y = x^2 + 8x + 12$$: The coefficient of $$x^2$$ is positive ($$+1$$), so the parabola opens upwards. The x-coordinate of the vertex is $$-b/(2a) = -8/(2*1) = -4$$. The y-coordinate of the vertex is $$(-4)^2 + 8(-4) + 12 = 16 - 32 + 12 = -4$$. So, the vertex is at $$(-4, -4)$$.
• Б) $$y = x^2 - 8x + 12$$: The coefficient of $$x^2$$ is positive ($$+1$$), so the parabola opens upwards. The x-coordinate of the vertex is $$-b/(2a) = -(-8)/(2*1) = 8/2 = 4$$. The y-coordinate of the vertex is $$(4)^2 - 8(4) + 12 = 16 - 32 + 12 = -4$$. So, the vertex is at $$(4, -4)$$.
• B) $$y = -x^2 + 8x - 12$$: The coefficient of $$x^2$$ is negative ($$-1$$), so the parabola opens downwards. The x-coordinate of the vertex is $$-b/(2a) = -8/(2*(-1)) = -8/(-2) = 4$$. The y-coordinate of the vertex is $$-(4)^2 + 8(4) - 12 = -16 + 32 - 12 = 4$$. So, the vertex is at $$(4, 4)$$.

Graph Analysis:

We need to visually inspect the graphs provided in the image to determine their characteristics.

• Graph labeled 1 (appears in the third row, first column): This graph shows a parabola opening upwards. The vertex appears to be in the third quadrant (negative x and negative y).
• Graph labeled 2 (appears in the third row, second column): This graph shows a parabola opening upwards. The vertex appears to be in the first quadrant (positive x and positive y).
• Graph labeled 3 (appears in the third row, third column): This graph shows a parabola opening downwards. The vertex appears to be in the first quadrant (positive x and positive y).

Matching:

Let's match the equations to the graphs based on our analysis.

• Equation A ($$y = x^2 + 8x + 12$$) has a vertex at $$(-4, -4)$$ and opens upwards. This vertex is in the third quadrant. This matches Graph 1.
• Equation Б ($$y = x^2 - 8x + 12$$) has a vertex at $$(4, -4)$$ and opens upwards. This vertex is in the fourth quadrant. However, among the upward-opening graphs with vertices in the first or fourth quadrant, this is the closest match for an upward opening graph with a vertex to the right of the y-axis. Graph 2 has its vertex in the first quadrant. If we consider the possibility of the vertex being in the first quadrant for Б, it is the most plausible upward opening graph to the right. Let's assume Graph 2 is the intended match for Б, despite the vertex location difference.
• Equation B ($$y = -x^2 + 8x - 12$$) has a vertex at $$(4, 4)$$ and opens downwards. This vertex is in the first quadrant. This matches Graph 3.

Therefore, the matches are: