I need help bad. Don't know which formula to use. Please explain step by step. 2x^2-12x+19. Find the interval where f is increasing and decreasing.
If you are confused by formulas, it indicates you don't really understand the concepts. Let's start from scratch.
You know the graph is a parabola, which has a vertex.
Since the coefficient of x^2 is positive, the vertex is at the minimum value of y.
So, left of the vertex y is decreasing, and on the right, it is increasing.
So, where is the vertex?
y = 2x^2 - 12x + 19
= 2(x^2-6x) + 19
= 2(x^2-6x+9) + 19 - 18
= 2(x-3)^2 + 1
So, the vertex is at (3,1)
y decreases for x < 3
y increases for x > 3
Vertex is -b/2a
ax^2+by+c=0
or
(x2-x1)/(y2-y1)
To determine the intervals where the function is increasing or decreasing, we need to find the critical points of the function.
Step 1: Find the derivative of the function.
Differentiate the given function, 2x^2 - 12x + 19, to find the derivative function, f'(x).
f'(x) = 4x - 12
Step 2: Set the derivative equal to zero and solve for x.
To find the critical points, we set f'(x) = 0 and solve for x.
4x - 12 = 0
4x = 12
x = 3
The value of x = 3 is the only critical point of the function.
Step 3: Determine the intervals.
Now we will analyze the intervals using the critical point.
1. If x < 3, substitute a value less than 3 into f'(x). For example, choose x = 0.
Plug x = 0 into f'(x).
f'(0) = 4(0) - 12
f'(0) = -12
Since f'(0) = -12 (negative), the function is decreasing on the interval x < 3.
2. If x > 3, substitute a value greater than 3 into f'(x). For example, choose x = 5.
Plug x = 5 into f'(x).
f'(5) = 4(5) - 12
f'(5) = 20 - 12
f'(5) = 8
Since f'(5) = 8 (positive), the function is increasing on the interval x > 3.
Therefore, the function is decreasing on the interval (-∞, 3) and increasing on the interval (3, +∞).