Question

answer the following questions about the quadratic function above. the graphs opens down. the vertex is (3, -4). this graph has a minimum. the y - intercept of the function is (3, -4).

Explanation:

1. Direction of Opening

Step1: Recall parabola opening rule

A parabola \( y = ax^2+bx + c \) opens up if \( a>0 \) (vertex is minimum) and down if \( a < 0 \) (vertex is maximum). From the graph, the parabola has a minimum point (since it opens up, wait the original selection was wrong). Wait, the graph shown has a U - shape, so it should open up. The vertex is the lowest point. So the correct direction: if the parabola has a minimum (vertex is lowest), it opens up. So the mistake in the original: the graph opens up, not down.

Step2: Correct the opening direction

Since the vertex is the minimum point (the lowest point of the parabola), the coefficient \( a>0 \), so the parabola opens up.

2. Vertex

Step1: Identify vertex from graph

The vertex is the point where the parabola changes direction. From the graph, the lowest point (vertex) has coordinates \( (3,-4) \), so this part was correct.

3. Minimum or Maximum

Step1: Relate vertex to min/max

If the parabola opens up (has a U - shape), the vertex is the minimum point (lowest point). If it opened down, it would be the maximum. Since the parabola opens up (correcting the first part), it has a minimum. Wait, but the graph's shape: looking at the graph, it's a U - shaped parabola (opening up), so the vertex is the minimum. But the original said "opens down" which was wrong. Let's re - evaluate:

Correcting each part:

1. Direction of Opening:

The parabola has a U - shape (the two ends go up), so it opens up. The rule is: for a quadratic function \( f(x)=a(x - h)^2 + k \), if \( a>0 \), opens up; \( a < 0 \), opens down. Since the vertex is the minimum, \( a>0 \), so opens up.

2. Vertex:

The vertex is the point \( (h,k) \) of the parabola. From the graph, the lowest point is at \( x = 3 \), \( y=-4 \), so vertex is \( (3,-4) \), correct.

3. Minimum/Maximum:

Since the parabola opens up, the vertex is the minimum point (the lowest point), so it has a minimum, this part was correct (even though the opening direction was wrong earlier, the fact that it has a minimum is correct because the vertex is the lowest point).

4. y - intercept:

Answer:

Step1: Recall y - intercept definition

The y - intercept is the point where \( x = 0 \) (the point where the parabola crosses the y - axis). From the graph, when \( x = 0 \), the parabola crosses the y - axis at \( (0,5) \) (looking at the grid, when \( x = 0 \), \( y = 5 \)). So the original selection \( (3,-4) \) is wrong. The y - intercept is found by plugging \( x = 0 \) into the function. If we assume the vertex form \( y=a(x - 3)^2-4 \), plug \( x = 0 \): \( y=a(0 - 3)^2-4=9a - 4 \). From the graph, when \( x = 0 \), \( y = 5 \), so \( 9a-4 = 5\Rightarrow9a=9\Rightarrow a = 1 \). So the function is \( y=(x - 3)^2-4=x^2-6x + 9-4=x^2-6x + 5 \). When \( x = 0 \), \( y = 5 \), so y - intercept is \( (0,5) \).

Final Answers:

1. The graph opens up.
2. The Vertex is \( (3,-4) \) (correct).
3. This graph has a minimum (correct, because it opens up, vertex is minimum).
4. The y - intercept of the function is \( (0,5) \).