For which k will the graph of f(x)=x^2−kx+k^2 cross the x-axis twice?
let's rephrase the question.
For what values of k will the quadratic
x^2 - kx + k^2 = 0 have 2 solutions
when b^2 - 4ac > 0
k^2 - 4(1)(k^2) >0
-3k^2 > 0
k^2 < 0
any number squared is always positive, so there is no value of k for which the graph will cross the x-axis twice
you can test this here
http://www.wolframalpha.com/input/?i=plot+y+%3D+x%5E2+-+5x+%2B+25
in my example I let k = 5
you can change the equation by editing the value of k, and the graph will always be above the x-axis
Well, my friend, let's solve this conundrum with a sprinkle of humor! To find the values of "k" where the graph of f(x) = x^2 - kx + k^2 crosses the x-axis twice, we need the discriminating factor. And like a discriminating eyebrow, this factor tells us how many solutions we've got.
The discriminant of a quadratic equation ax^2 + bx + c = 0 is given by b^2 - 4ac. In our case, a is 1, b is -k, and c is k^2. So, the discriminant is (-k)^2 - 4(1)(k^2), which simplifies to k^2 - 4k^2.
Now, for the graph to cross the x-axis twice, the equation must have two distinct real solutions. In other words, the discriminant must be greater than zero, like a double scoop of ice cream on a sunny day. So, we have:
k^2 - 4k^2 > 0
Simplifying this inequality, we get -3k^2 > 0.
To find the range of "k" that makes this true, we need to consider whether the parabola opens upward or downward. Since the coefficient of x^2 is positive (1), the parabola opens upward. Hence, the entire range of k makes the graph cross the x-axis twice!
So, my friend, the graph of f(x)=x^2−kx+k^2 will have two x-intercepts for all values of "k". It's like having an endless supply of laughter—there's no shortage of solutions here!
To determine the values of k for which the graph of f(x) = x^2 - kx + k^2 will cross the x-axis twice, we need to find the discriminant of the quadratic equation formed by setting f(x) equal to zero.
The quadratic equation is x^2 - kx + k^2 = 0.
The discriminant (D) of a quadratic equation ax^2 + bx + c = 0 is given by D = b^2 - 4ac.
In this case, a = 1, b = -k, and c = k^2. Substituting these values into the formula for D, we have:
D = (-k)^2 - 4(1)(k^2)
D = k^2 - 4k^2
D = -3k^2
For the graph of the quadratic equation to cross the x-axis twice, the discriminant D must be greater than zero (D > 0).
Therefore, we have:
-3k^2 > 0
To solve this inequality, we divide both sides of the inequality by -3, but since we are dividing by a negative number, the inequality sign flips:
k^2 < 0
Taking the square root of both sides of the inequality gives:
k < 0
So, for the graph of f(x) = x^2 - kx + k^2 to cross the x-axis twice, the value of k must be less than 0.
To determine for which values of k the graph of f(x) = x^2 - kx + k^2 will cross the x-axis twice, we need to consider the discriminant of the quadratic equation.
The discriminant (D) of a quadratic equation, ax^2 + bx + c = 0, is given by D = b^2 - 4ac. For a quadratic equation to have two distinct real roots (cross the x-axis twice), the discriminant must be positive (D > 0).
In this case, we have f(x) = x^2 - kx + k^2, so the coefficients are a = 1, b = -k, and c = k^2. Substituting these values into the discriminant formula, we get:
D = (-k)^2 - 4(1)(k^2)
= k^2 - 4k^2
= -3k^2
For the graph to cross the x-axis twice, the discriminant must be positive, so -3k^2 > 0. To solve this inequality for k, we divide both sides of the inequality by -3, remembering to reverse the inequality direction because we are dividing by a negative number:
k^2 < 0
Since k^2 is always positive or zero (since it's a square), there is no solution to the inequality k^2 < 0. Therefore, there is no value of k for which the graph of f(x) = x^2 - kx + k^2 will cross the x-axis twice.