f(x)=4x^2-8x+9
find the vertex the maximum the minimum the range and the intervals where it increases and decreases
Try completing the square first.
4x^2 -8x +9 = 4(x^2 -2x +1) +9 -4
= 4(x-1)^2 + 5
The vertex is where it has its lowest value, which is obviously at x=1, f(x) = 5
The function increases for x<1 and x>1
There is no maximum.
The range is 5 to infinity
To find the vertex, maximum, minimum, range, and intervals of increase and decrease for the given function f(x) = 4x^2 - 8x + 9, we can follow these steps:
Step 1: Find the vertex:
- The vertex of a quadratic function can be found using the formula x = -b / (2a), where a, b, and c are the coefficients of the quadratic equation ax^2 + bx + c.
In this case, a = 4 and b = -8. Replacing the values in the formula, we get:
x = -(-8) / (2 * 4)
x = 8 / 8
x = 1
To find the y-coordinate of the vertex, substitute x = 1 into the given equation:
f(1) = 4(1)^2 - 8(1) + 9
f(1) = 4 - 8 + 9
f(1) = 5
Therefore, the vertex is (1, 5).
Step 2: Determine whether the vertex is a maximum or minimum:
- Since the coefficient of the x^2 term (a) is positive (+4), the parabola opens upwards, indicating that the vertex is a minimum.
Step 3: Find the maximum/minimum value:
- The maximum or minimum value of a quadratic function occurs at the vertex, so the y-coordinate of the vertex, which is 5, represents the minimum value.
Therefore, the minimum value of the function is 5.
Step 4: Determine the range:
- Since the vertex represents the minimum value, the range will be all real numbers greater than or equal to the y-coordinate of the vertex, which is 5.
Therefore, the range of the function is [5, ∞).
Step 5: Identify the intervals of increase and decrease:
- A quadratic function can either increase or decrease between turning points, which means we need to find the x-values where f'(x) = 0 or f'(x) does not exist, where f'(x) represents the derivative of f(x).
Taking the derivative of f(x) = 4x^2 - 8x + 9, we get:
f'(x) = 8x - 8
Setting f'(x) = 0, we have:
8x - 8 = 0
8x = 8
x = 1
By solving this equation, we find that the function has a critical point at x = 1.
Now we can determine the intervals of increase/decrease by considering intervals on either side of this critical point.
- For x < 1, we can choose x = 0 as a test point. Substituting x = 0 into f'(x), we get:
f'(0) = 8(0) - 8
f'(0) = -8
Since f'(x) is negative, the function is decreasing for x < 1.
- For x > 1, we can choose x = 2 as a test point. Substituting x = 2 into f'(x), we get:
f'(2) = 8(2) - 8
f'(2) = 8
Since f'(x) is positive, the function is increasing for x > 1.
Therefore, the function decreases for x < 1 and increases for x > 1.
To summarize:
- The vertex is (1, 5).
- The vertex represents the minimum.
- The minimum value of the function is 5.
- The range of the function is [5, ∞).
- The function decreases for x < 1 and increases for x > 1.