The question is:
Fr what value(s) of k will the function f(x)=(kx^2)-4x+k have no zeroes.
So for that, the discriminant would be less than 0.
0>D
0>((-4)^2)-4(k)(k)
0>16-(4k^2)
4k^2>16
k^2>4
k>+2/-2
So I know that I'm supposed to graph it with D as one axis an k as the other, and then shade the part of the graph that agrees with D<0, and use that to make the restrictions.
BUT I don't know whether the graph hs a minimum or maximum because I don't know the k value which is the a value that determines whether or not the graph has a minimum or maximum.
How do I finish this question?
Thanks.
To determine whether the graph of the function has a minimum or maximum, you can look at the coefficient of the x^2 term in the function f(x). In this case, the coefficient is k.
If k is positive, the graph will open upwards and have a minimum point.
If k is negative, the graph will open downwards and have a maximum point.
If k is zero, the graph will be a straight line.
Since you are looking for values of k that make the function have no zeroes, you want the graph to either always be above the x-axis or always be below the x-axis.
Considering k = 2 and k = -2 separately:
For k = 2: The function becomes f(x) = 2x^2 - 4x + 2.
To find the discriminant, you can use the formula: D = b^2 - 4ac. In this case, a = 2, b = -4, and c = 2.
D = (-4)^2 - 4(2)(2) = 16 - 16 = 0. This means that the graph of the function will touch the x-axis at exactly one point, so it will have a double root.
For k = -2: The function becomes f(x) = -2x^2 -4x -2.
Again, we can find the discriminant using the formula: D = b^2 - 4ac. In this case, a = -2, b = -4, and c = -2.
D = (-4)^2 - 4(-2)(-2) = 16 - 16 = 0. Like before, the graph of the function will touch the x-axis at exactly one point, so it will have a double root.
Based on the above calculations, it appears that there are no values of k that will make the function have no zeroes. The function will always have at least one zero (a double root) when k is either 2 or -2.