Find the maximum or minimum of the following quadratic function: y = x2 - 2x - 4?

Question

i dont understand how to do this if someone can work it out and explain as they do so maybe i can get it...

Answer 1

For a quadratic equation, the maximum or minimum will be the vertex. Since the first term of the quadratic is positive, the parabola goes up, so you're looking for the minimum.

There are 2 methods:
Calculus (using derivatives):
f(x) = x2 - 2x - 4
f'(x) = 2x - 2
the maximum/minimum will be where f'(x) = 0.
2x - 2 = 0
x = 1
y = 1^2 - 2*1 - 4 = -5

Algebra:
The vertex can be found using the formula:
(-b/2a , -(b^2 - 4ac)/4a)
(-(-2)/2, -(4 - (4*1*-4))/4)
(1,-5)

Check: x = 1, y = 1^2 - 2*1 - 4 = -5

Answer 2

There are several methods of finding Max or Min of a function. i will list the popular first 3 for you;
i) derivatives method
ii) Graphical method
iii) Vertex method
Now due tothis medium of communication i will use method (i).

here are the conditions in summary;
The function has a minimum value at x = a if f '(a) = 0
and f ''(a) = a positive number.

The function has a maximum value at x = a if f '(a) = 0
and f ''(a) = a negative number.
where f' and f'' are first and second derivatives respectively.
x = a is the turning point or critical value. ( remember to find the corresponding value of y at x=a)

example1
Example 2. Let f(x) = 2x^3 − 9x² + 12x − 3.

Are there any extreme values? That is, are there any critical values, and do they determine a maximum or a minimum? And what are the coordinates on the graph of that maximum or minimum -- where are the turning points?

Solution. f '(x) = 6x² − 18x + 12 = 6(x² − 3x + 2)

= 6(x − 1)(x − 2) = 0

implies: x = 1 or x = 2.

These are the critical values. Does each one determine a maximum or a minimum? To answer, we must evaluate the second derivative at each value.
f '(x) = 6x² − 18x + 12.
let do x = 1 first
f ''(x) = 12x − 18.
f ''(1) = 12 − 18 = −6.

The second derivative is negative. The function therefore has a maximum at x = 1.

To find the y-coordinate -- the extreme value -- at that maximum we evaluate f(1):
f(x) = 2x3 − 9x² + 12x − 3
f(1) = 2 − 9 + 12 − 3 = 2.
The maximum occurs at the point (1, 2).

Next, does x = 2 determine a maximum or a minimum?
f ''(x) = 12x − 18.
f ''(2) = 24 − 18 = 6.
The second derivative is positive. The function therefore has a minimum at x = 2.
To find the y-coordinate -- the extreme value -- at that minimum, we evaluate f(2):
f(x) = 2x3 − 9x² + 12x − 3.
f(2) = 16 − 36 + 24 − 3 = 1.
The minimum occurs at the point (2, 1).

Now to your question;

f(x) = x^2 - 2x - 4
the first derivative will give
f'(x) = 2x - 2
the critical points(values) will be at f'(x) = 0.
2x - 2 = 0
x = 1 -----this is the critical value, now find the corresponding y
y = 1^2 - 2*1 - 4 = -5
( x,y) = (1, -5) is this Max or Min? Let us find out.

recall that;
The function has a minimum value at x = a if f '(a) = 0
and f ''(a) = a positive number.

The function has a maximum value at x = a if f '(a) = 0
and f ''(a) = a negative number.

f'(x) = 2x-2
and f''(x) = 2 ------a positive number >>>>>this satisfy the first condition, Therefore (1,-5) is a minimum point.

Answer 3

this would not element over the rationals. Use the quadratic formulation to discover x intercepts: -2 +- sqrt(4 - 28) / -2 (no x intercepts, ideas are non-genuine) the y intercept is -7 (C is your y intercept because of the fact the y-intercept is the place x is comparable to 0, subbing a nil in for x leaves in basic terms C). optimal element is (-b/2a, f(-b/2a): (a million, -6) Therfore the optimal fee is -6.