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The entire graphs of the equation y = x2 + kx – x + 9 is strictly above the x-axis if and only if
- a)k < 7
- b)– 5 < k < 7
- c)k > – 5
- d)None of these.
Correct answer is option 'B'. Can you explain this answer?
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The entire graphs of the equation y = x2 + kx – x + 9 is strictl...
For entire graph to be above x-axis, we should have
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The entire graphs of the equation y = x2 + kx – x + 9 is strictl...
We cannot provide the full graph of the equation y = x^2 + kx as it depends on the value of "k" which is not specified in the question. However, we can discuss the general shape and behavior of the graph.
The equation y = x^2 + kx represents a quadratic function, which is a curve in the shape of a parabola. The vertex of the parabola is at the point (-k/2, -k^2/4), which is the minimum or maximum point of the function depending on the value of "k".
If "k" is positive, then the parabola opens upwards and the vertex is a minimum point. The graph will be symmetric with respect to the vertical line passing through the vertex. As x moves further away from the vertex, the value of y increases rapidly.
If "k" is negative, then the parabola opens downwards and the vertex is a maximum point. The graph will also be symmetric with respect to the vertical line passing through the vertex. As x moves further away from the vertex, the value of y decreases rapidly.
In general, the graph of y = x^2 + kx will intersect the x-axis at two points if k is positive, and will not intersect the x-axis if k is negative. It will always intersect the y-axis at the point (0,0).
Without knowing the value of "k", we cannot provide a specific graph, but the general shape and behavior of the parabola can be predicted based on the sign of "k".
The equation y = x^2 + kx represents a quadratic function, which is a curve in the shape of a parabola. The vertex of the parabola is at the point (-k/2, -k^2/4), which is the minimum or maximum point of the function depending on the value of "k".
If "k" is positive, then the parabola opens upwards and the vertex is a minimum point. The graph will be symmetric with respect to the vertical line passing through the vertex. As x moves further away from the vertex, the value of y increases rapidly.
If "k" is negative, then the parabola opens downwards and the vertex is a maximum point. The graph will also be symmetric with respect to the vertical line passing through the vertex. As x moves further away from the vertex, the value of y decreases rapidly.
In general, the graph of y = x^2 + kx will intersect the x-axis at two points if k is positive, and will not intersect the x-axis if k is negative. It will always intersect the y-axis at the point (0,0).
Without knowing the value of "k", we cannot provide a specific graph, but the general shape and behavior of the parabola can be predicted based on the sign of "k".
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The entire graphs of the equation y = x2 + kx – x + 9 is strictly above the x-axis if and only ifa)k < 7b)– 5 < k < 7c)k > – 5d)None of these.Correct answer is option 'B'. Can you explain this answer?
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