maximum/minimum of quadratic function?

Looking at the leading coefficient (the number in front of x^2) we see that it is positive. That means that the graph is a parabola opening upwards, so this function has a minimum.

The x-coordinate of the vertex is found using the formula x = -b/(2a)

so, x = -8/2 = -4. The y - coordinate is (-4)^2 + 8*(-4) + 10 = -6.

So, the minimum value is -6 and it occurs when x = -4.

There are two good ways to do this.

Graphically we can plot this function on a TI calculator then use the built in min and max functions to give us the coordinates.

(The min and max functions are under GRAPH then MATH)

However if you know Calculus you can also use the derivative.

For more info on Minimums and Maximums

look up these topics

"Extrema"

http://en.wikipedia.org/wiki/Maxima_and_minima

"Critical Points"

http://en.wikipedia.org/wiki/Critical_point_(mathe...

These occur when the derivative of a function = 0.

f ' (x) = 0

Thus first find the derivative, then solve for when x = 0.

f ' (x) = 2x + 8

f ' (x) = 0

0 = 2x + 8

0 = 2( x + 4)

x = -4

This tells me that one extrema is at x = -4

Now plug that x value into our function to get our y min or max.

f(-4) = (-4)² + 8(-4) + 10

= 16 - 32 + 10

= -16 + 10

= -6

So we have an extrema at (-4, -6

This happens to be the minimum of this function.

The function looks like a u so there is only one extrema.

By the way x² functions always look u-shaped.

Hi,

You can find the maximum/minimum of a quadratic equation by the concept of differentiation.

By the concept of MAXIMA $ MINIMA we get the concept .

f(x)=x^2+8x+10

f'(x)=2x+8

f''(x)=2>0

By double differentiatiate we come to know that the function has minimum at f'(x)=0

2x+8=0 (As f'(x)=0 gives the maximum or minimum value)

x= --4.

So we find f(--4)=(--4)^2+8(--4)+10

=16--32+10 = --6

Therefore the minimum value of the function is --6.

3 steps:

1. Calculate the 1st & 2nd derivatives of f(x) with respect to x.

df/dx = 2x + 8

d^2f/dx^2 = 2

2. Set the 1st derivative equal to 0, and solve for x.

0 = 2x + 8

x = -4

You know that the max, min, or inflection point occurs at x = -4,

3. To determine which of the 3 it is, find the value of the 2nd derivative at x = -4. A positive value means it is a minimum; negative means maximum; and zero means inflection point.

Since the 2nd derivative is constant, equal to 2, this point is a minimum.

So, at x = -4, f(x) = (-4)^2 + 8*(-4) + 10 = -6

Check some other point, like x = -5, to see that f(-5) = 25 -40 + 10 = -5

So, we have confirmed that x = -4 is a minimum, with f(x) = -6.

hi, f(x) = x(x-2) let's multiply this and we've. f(x) = x^2 - 2x. Now for a regular quadratic ax^2 + bx + c then the x-fee of the vertex the position both the max or min occurs is stumbled on via creating use of x = -b/(2*a) so for this one we've a=a million, b=-2 and c = 0. Then x = -(-2)/(2*a million) for this reason x = a million now to discover the y-fee or min fee in our case we positioned x = a million into the function therefore f(a million) = a million(a million-2) = -a million. The vertex is at (a million,-a million) and is a min because the a fee is efficacious the graph opens upward. wish This helps!!

df(x)/dx = 2*x +8

For Max/Min df(x)/dx = 0

2*x +8 = 0 -------> x = - 4

no max.

min of f is

x=-b/2a

a=1

b=8

c=10

x= - 4

y= - 6

(-4, -6) min point

Tell me what that little ^ sign means and I can help you.