Answer:
No answer is possible
Step-by-step explanation:
First, we can identify what the parabola looks like.
A parabola of form ax²+bx+c opens upward if a > 0 and downward if a < 0. The a is what the x² is multiplied by, and in this case, it is positive 2. Therefore, this parabola opens upward.
Next, the vertex of a parabola is equal to -b/(2a). Here, b (what x is multiplied by) is 1 and a =2, so -b/(2a) = -1/4 = -0.25.
This means that the parabola opens upward, and is going down until it reaches the vertex of x=-0.25 and up after that point. Graphing the function confirms this.
Given these, we can then solve for when the endpoints of the interval are reached and go from there.
The first endpoint in -2 ≤ f(x) ≤ 16 is f(x) = 2. Therefore, we can solve for f(x)=-2 by saying
2x²+x-4 = -2
add 2 to both sides to put everything on one side into a quadratic formula
2x²+x-2 = 0
To factor this, we first can identify, in ax²+bx+c, that a=2, b=1, and c=-2. We must find two values that add up to b=1 and multiply to c*a = -2 * 2 = -4. As (2,-2), (4,-1), and (-1,4) are the only integer values that multiply to -4, this will not work. We must apply the quadratic formula, so
x= (-b ± √(b²-4ac))/(2a)
x = (-1 ± √(1-(-4*2*2)))/(2*2)
= (-1 ± √(1+16))/4
= (-1 ± √17) / 4
when f(x) = -2
Next, we can solve for when f(x) = 16
2x²+x-4 = 16
subtract 16 from both sides to make this a quadratic equation
2x²+x-20 = 0
To factor, we must find two values that multiply to -40 and add up to 1. Nothing seems to work here in terms of whole numbers, so we can apply the quadratic formula, so
x = (-1 ± √(1-(-20*2*4)))/(2*2)
= (-1 ± √(1+160))/4
= (-1 ± √161)/4
Our two values of f(x) = -2 are (-1 ± √17) / 4 and our two values of f(x) = 16 are (-1 ± √161)/4 . Our vertex is at x=-0.25, so all values less than that are going down and all values greater than that are going up. We can notice that
(-1 - √17)/4 ≈ -1.3 and (-1-√161)/4 ≈ -3.4 are less than that value, while (-1+√17)/4 ≈ 0.8 and (-1+√161)/4 ≈ 2.9 are greater than that value. This means that when −2 ≤ f(x) ≤ 16 , we have two ranges -- from -3.4 to -1.3 and from 0.8 to 2.9 . Between -1.3 and 0.8, the function goes down then up, with all values less than f(x)=-2. Below -3.4 and above 2.9, all values are greater than f(x) = 16. One thing we can notice is that both ranges have a difference of approximately 2.1 between its high and low x values. The question asks for a value of a where a ≤ x ≤ a+3. As the difference between the high and low values are only 2.1, it would be impossible to have a range of greater than that.
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