The solutions of the considered function h(x) are x = -1 and x = 2. The nature of solutions is given by: Option B: All real solutions.
Suppose that we've a function y = f(x) such that f(x) is quadratic.
When y = 0, then the values of x for which f(x) = 0 is called solution of quadratic equation f(x) = 0
These solution gives values of x, and when we plot x and f(x), we'd see that the graph intersects the x-axis at its solution points.
Example: For x² + 5x + 6 = 0, the solutions are x = -3, and x = -2. The graph of y = x² + 5x + 6 will intersect the x-axis at x= -3 and x=-2 (and those are the only intersection it will do on x-axis).
For this case, the graph of y = h(x) intersects the x-axis at x = -1 and x = 2
Thus, the solution of h(x) = 0 is x = -1 and x = 2 (this shows that if we put x=-1 or x=2 in h(x), we will get h(x) = 0) ).
As -1 is not a whole number of positive, and neither all the points lying on graph of y = h(x) is its solution, so the option B is the only correct option as -1 and 2 both are real.
Thus, the solutions of the considered function h(x) are x = -1 and x = 2. The nature of solutions is given by: Option B: All real solutions.
Learn more about solutions of a quadratic equation here:
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