Create a quadratic function in which f(x) ≥ 0 for values of x between 2 and 6 inclusive.
if the parabola opens downward, it is positive between the roots. Draw one and see. So, for any negative value of a,
y = a(x-2)(x-6) ≥ 0 on the interval [2,6]
@oobleck can you clarify your answer please. How can there be both an equal sign and a comparison sign in one equation?
To create a quadratic function that satisfies the condition f(x) ≥ 0 for values of x between 2 and 6 inclusive, we can follow these steps:
Step 1: Determine the general form of a quadratic function
The general form of a quadratic function is f(x) = ax^2 + bx + c, where a, b, and c are constants.
Step 2: Determine the vertex of the quadratic function
For the function to be positive between 2 and 6, the vertex of the parabola should lie within this range. The x-coordinate of the vertex can be determined using the formula x = -b / (2a). Since the vertex lies between 2 and 6, we can set 2 ≤ x ≤ 6.
Step 3: Choose the value of "a"
To ensure the quadratic function opens upwards and maintains a positive value within the given range, we should choose a positive value for "a". Let's consider a = 1 for simplicity.
Step 4: Determine the x-intercepts (roots) of the quadratic function
The x-intercepts occur when f(x) = 0. Since f(x) ≥ 0 for values of x between 2 and 6, there should be no real roots in this interval.
Step 5: Determine the coefficients "b" and "c"
To further determine the function, we need to find the values of "b" and "c". Since we already have a value for "a" (a = 1), we can solve for the quadratic equation f(x) = x^2 + bx + c using specific points.
Step 6: Substitute the vertex (x, y) values
Since we know the vertex lies within the interval 2 ≤ x ≤ 6, we can substitute the value into the equation to solve for c.
Let's proceed with the calculations:
Step 1: f(x) = ax^2 + bx + c
Step 2: x = -b / (2a), 2 ≤ x ≤ 6
-2 / a ≤ b ≤ -6 / a
Step 3: Let a = 1 (to ensure the quadratic function opens upwards)
Step 4: No real roots should exist in the interval 2 ≤ x ≤ 6
Step 5: f(x) = x^2 + bx + c
Plug in points (2, 0) and (6, 0) since f(x) ≥ 0 for values of x between 2 and 6 inclusive.
When x = 2: 2^2 + 2b + c = 0
When x = 6: 6^2 + 6b + c = 0
Step 6: Determine the value of c using the vertex
Since the vertex lies within the interval 2 ≤ x ≤ 6, substitute the x-coordinate of the vertex, which is given by x = -b / (2a), into the equation to solve for c.
Once these steps are followed, you should have a quadratic function that meets the given conditions.
To create a quadratic function in which f(x) ≥ 0 for values of x between 2 and 6 inclusive, we can start by considering the properties of a quadratic function in vertex form.
A quadratic function in vertex form is given by the equation:
f(x) = a(x - h)^2 + k
Where (h, k) represents the vertex of the parabola. In order to ensure that the function is greater than or equal to zero for values of x between 2 and 6, we want the vertex (h, k) to lie within that interval.
Let's set h = 4, which is the midpoint of the interval [2, 6].
Thus, we have:
f(x) = a(x - 4)^2 + k
Now, let's determine the value of a. To ensure that f(x) is always greater than or equal to zero, the coefficient 'a' should be positive. Let's choose a = 1 for simplicity. This means we have:
f(x) = (x - 4)^2 + k
Since we want f(x) ≥ 0 for values of x between 2 and 6 inclusive, we can substitute these x-values into the equation and solve for k.
For x = 2:
0 ≤ (2 - 4)^2 + k
Simplifying:
0 ≤ 4 + k
Since k can be any value to satisfy f(x) ≥ 0, we can choose k = -4. This means our quadratic function becomes:
f(x) = (x - 4)^2 - 4
Therefore, the quadratic function f(x) = (x - 4)^2 - 4 will be greater than or equal to zero for values of x between 2 and 6 inclusive.