Does the graph of y = x – 3x2 + 5 have a maximum or minimum?
What is the vertex of the graph of y = -2(x - 3)2 + 4?
Does the graph of y = -2(x - 3)2+ 4 open up or down?
I will be happy to critique your thinking on these; Post your thoughts on each one.
If you are currently studying the quadratic function and its graph, then these are fundamental questions that you should know.
If you are not studying the quadratic function then these pages are an excellent source
http://www.uncwil.edu/courses/mat111hb/Pandr/quadratic/quadratic.html
http://www.analyzemath.com/quadraticg/quadraticg.htm
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To determine whether the graph has a maximum or minimum, we need to consider the coefficient of the x^2 term, which is -3 in the equation y = x – 3x^2 + 5.
In general, when the coefficient of the x^2 term in a quadratic function is negative, the graph opens downward, which means it will have a maximum value. Conversely, if the coefficient is positive, the graph opens upward, which means it will have a minimum value.
So, since the coefficient of x^2 in the given equation is -3 (negative), the graph will have a maximum value.
To find the vertex of the graph of y = -2(x - 3)^2 + 4, we can use the vertex form of a quadratic function, which is y = a(x - h)^2 + k. Here, (h, k) represents the coordinates of the vertex.
In this case, we have y = -2(x - 3)^2 + 4. Comparing this with the vertex form, we can see that h = 3 and k = 4. Therefore, the vertex of the graph is (3, 4).
To determine if the graph of y = -2(x - 3)^2 + 4 opens up or down, we again consider the coefficient of the x^2 term, which is -2 in this equation.
Similarly to the previous explanation, when the coefficient of the x^2 term is negative, the graph opens downward. Conversely, when it is positive, the graph opens upward.
Since the coefficient of x^2 in y = -2(x - 3)^2 + 4 is -2 (negative), the graph opens downward.