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Why does the graph of an absolute-value function not extend past the vertex?
The graph of an absolute-value function does not extend past the vertex because the vertex represents the minimum (or maximum, in the case of a downward-opening parabola) point of the graph. The absolute value ensures that all output values are non-negative (or non-positive), meaning that as you move away from the vertex in either direction, the values will either increase or decrease but never go below the vertex value. Consequently, the graph is V-shaped and reflects this property, making it impossible for the graph to extend beyond the vertex in the negative direction.
What is the answer to graph the function f(x) x2 4?
To graph the function ( f(x) = x^2 + 4 ), recognize that it is a quadratic function with a vertex at (0, 4). The parabola opens upwards, and the y-intercept is at (0, 4). As ( x ) increases or decreases from the vertex, the values of ( f(x) ) will rise, creating a symmetric shape around the y-axis. To sketch the graph, plot the vertex and a few additional points, and then draw the parabola.
What is a cycle in graph theory?
If the graph start and end with same vertex and no other vertex can be repeated then it is called trivial graph.
Which way does the graph open when 'a' is positive?
When 'a' is positive in a quadratic function of the form (y = ax^2 + bx + c), the graph opens upwards. This means the vertex of the parabola is the lowest point on the graph, and as you move away from the vertex in either direction, the values of (y) increase.
How do you graph parabola given its vertex?
The vertex of a parabola doe not provide enough information to graph anything - other than the vertex!
Which vertex in the graph does not have any weighting assigned to it?
The vertex that does not have any weighting assigned to it in the graph is called an unweighted vertex.
What name is given to the turning point also known as the maximum or minimum of the graph of a quadratic function?
vertex
What different information do you get from vertex form and quadratic equation in standard form?
The graph of a quadratic function is always a parabola. If you put the equation (or function) into vertex form, you can read off the coordinates of the vertex, and you know the shape and orientation (up/down) of the parabola.
What is a cycle in graph theory?
If the graph start and end with same vertex and no other vertex can be repeated then it is called trivial graph.
How do you graph parabola given its vertex?
The vertex of a parabola doe not provide enough information to graph anything - other than the vertex!
What point on the graph could be the fourth vertex of the parallelogram?
the origin is the point in the graph that can be fourth vertex
Is x2 plus 8 a function?
x2+8= y This equation represents a function. It will be a parabola with the vertex at (0,8). You can easily graph this on a graphing calculator or from prior knowledge. You know the basic graph of y=x2 with vertex (0,0) and opens upwards on the y-axis. From the equation, you simply shift the vertex vertically up 8 so the new vertex is (0,8) This represents a function because for every x value there is one y value.
What is the highest or lowest point on a graph called?
The vertex is the highest or lowest point on a graph.
What is an example of range in math?
range is the y values in a graph otherwise known as a function; for example in the graph y= abs(x), the graph is a v with the vertex at the origin and the range is (0,infinity).
What is y equals x2 plus 2x plus 1 on a graph?
Interpreting that function as y=x2+2x+1, the graph of this function would be a parabola that opens upward. It would be equivalent to y=(x+1)2. Its vertex would be at (-1,0) and this vertex would be the parabola's only zero.
What is a pendent vertex?
A vertex of a graph is said to be pendant if its neighborhood contains exactly one vertex.
What is the complexity of finding the minimum vertex cover in a graph, also known as the vertex cover problem?
The complexity of finding the minimum vertex cover in a graph, also known as the vertex cover problem, is NP-hard.
