Quadratic function written in standard form where a, b, and c are constants such that a is not zero.
f(x)= ax^2+bx+c
Using calculus find the vertex of the parabola formed by this quadratic function. Determine under what conditions this vertex is a maximum or minumum (using Calculus techniques). Show work using derivatives to justify our conclusions.
THis is what I did so far, but can't figure out the rest.
f'(x)= 2ax + b
0=2ax+b
-b = 2ax
-b/2a = x
Please help and show steps so that I can understand.
That is correct. The minimum ofr maximum (vertex) is at x = -b/2a
You already know that from the quadratic equation. x = -b/2a +/- (b/2a)sqrt (b^2-4ac)
Now
if x = -b/2a
what is y?
y = a (b^2/4a^2) + b(-b/2a) + c
= b^2/4a -b^2/2a + c
= -b^2/4a+c
for max or min
f" = 2 a
if a is +, that is a minimum
if a is -, that is a maximum
To find the vertex of the parabola formed by the quadratic function, we need to find the x-coordinate of the vertex. The x-coordinate of the vertex can be found using calculus by finding the derivative of the quadratic function and setting it equal to zero.
You've correctly taken the derivative of the quadratic function, which is f'(x) = 2ax + b. To find the x-coordinate of the vertex, we set the derivative equal to zero:
2ax + b = 0
To isolate x, we can subtract b from both sides:
2ax = -b
Next, divide both sides by 2a:
x = -b/2a
Now we have the x-coordinate of the vertex, which is -b/2a.
To determine whether the vertex is a maximum or minimum, we need to analyze the concavity of the parabola using the second derivative. The second derivative of the quadratic function can be found by taking the derivative of the first derivative:
f''(x) = 2a
Since the second derivative is constant, it does not change with x. If the second derivative (2a) is positive, the graph is concave up, and the vertex represents a minimum point. If the second derivative is negative, the graph is concave down, and the vertex represents a maximum point.
So, in summary:
- The x-coordinate of the vertex is x = -b/2a.
- If the second derivative f''(x) = 2a is positive, the vertex is a minimum.
- If the second derivative is negative, the vertex is a maximum.
By analyzing the sign of the second derivative, you can determine whether the vertex represents a maximum or minimum point on the parabola.