Which is the graph of y = 2(x – 3)2 + 2
(1 point)
Responses
graphThe function has a maximum at left-parenthesis 3 comma 2 right-parenthesis and passes through the points left-parenthesis 2 comma 0 right-parenthesis and left-parenthesis 4 comma 0 right-parenthesis.
Image with alt text: graph The function has a maximum at left-parenthesis 3 comma 2 right-parenthesis and passes through the points left-parenthesis 2 comma 0 right-parenthesis and left-parenthesis 4 comma 0 right-parenthesis.
graphThe function has a minimum at left-parenthesis 3 comma 2 right-parenthesis and passes through the points left-parenthesis 2 comma 4 right-parenthesis and left-parenthesis 4 comma 4 right-parenthesis.
Image with alt text: graph The function has a minimum at left-parenthesis 3 comma 2 right-parenthesis and passes through the points left-parenthesis 2 comma 4 right-parenthesis and left-parenthesis 4 comma 4 right-parenthesis.
graphThe function has a minimum at left-parenthesis negative 3 comma 2 right-parenthesis and passes through the points left-parenthesis negative 4 comma 4 right-parenthesis and left-parenthesis negative 2 comma 4 right-parenthesis.
Image with alt text: graph The function has a minimum at left-parenthesis negative 3 comma 2 right-parenthesis and passes through the points left-parenthesis negative 4 comma 4 right-parenthesis and left-parenthesis negative 2 comma 4 right-parenthesis.
graph
The function has a minimum at (3, 2) and passes through the points (2, 4) and (4, 4).
The graph of the equation y = 2(x - 3)^2 + 2 is a parabola with its vertex at (3, 2). It opens upward because the coefficient of the squared term, 2, is positive. The parabola passes through the points (2, 0) and (4, 0).
To graph the function y = 2(x – 3)2 + 2, we can follow these steps:
1. Identify the vertex of the parabola. The equation is in the form y = a(x-h)2 + k, where the vertex is at the point (h, k). In this case, the vertex is (3, 2).
2. Plot the vertex on the coordinate plane.
3. Determine if the parabola opens upward or downward. Since the coefficient of the x-term is positive (2), the parabola opens upward.
4. Determine the points where the parabola intersects the x-axis. To find these points, set y = 0 and solve for x. In this case, we have 0 = 2(x – 3)2 + 2. Solving this equation gives (x - 3)2 = -1, which has no real solutions. Therefore, the parabola does not intersect the x-axis.
5. Sketch the graph of the parabola symmetrically around the vertex, keeping in mind the direction in which it opens.
Based on the above steps, the correct graph would be the one that says "graphThe function has a minimum at (3, 2) and passes through the points (2, 4) and (4, 4)."