Question

For f(x)=a(x-h)^2+k, identify which constant (a, h or k) represents the max or min value of the function.

I'm only supposed to put down 1 answer, but I thought h or k both could determine the max or min? This one just got me a lil confused

Answer 1

The a represents the amplitude but at the same time it also represents if the function will have a maximum or a minimum. So if a is negative we will have a maximum and if a is positive we will have a minimum.

ohhhh actually I was thinking that it may be the k which is the height of the maximum or minimum value. It doesnt ask you for the coordinates of the max or min it asks you for the max or min which is the y - coordinate of the max or min which is k

Answer 2

@mightylions1 Ya the more i think abt it the more i am convinced that its k

Answer 3

My view is different: I believe that locating the point that represents the maximum of this function is most important, and that of the two coordinates that represent that point, the x-coordinate is the more important one.

If and when you take calculus, you'll learn that calculus methods of finding the min. or max. always produce the x-coordinate (dubbed "critical value," not the y-coordinate.

So: what is the x-coordinate in this particular problem?

Answer 4

@BlackLabel interesting point you brought up, a does control the direction, so it would affect the max or min. But then @Mathmale does have a good point, the max or min will always be a point up or down on the x axis... so then it would be h. Tough one but h might be the answer. Anyone else have ideas on this?

Answer 5

What do u need help with

Answer 6

@geekfromthefutur read the question....

Answer 7

ok

Answer 8

if i help you @mightylions1 u help me ok

Answer 9

tThe term (x - h)2 is a square, hence is either positive or equal to zero.

(x - h)2 ≥ 0

If you multiply both sides of the above inequality by coefficient a, there are two possibilities to consider, a is positive or a is negative.

case 1: a is positive
a(x - h)2 ≥ 0.

Add k to the left and right sides of the inequality

a(x - h)2 + k ≥ k.

The left side represents f(x), hence f(x) ≥ k. This means that k is the minimum value of function f.

case 2: a is negative
a(x - h)2 ≤ 0.

Add k to the left and right sides of the inequality

a(x - h)2 + k ≤ k.

The left side represents f(x), hence f(x) ≤ k. This means that k is the maximum value of function f.

Note also that k = f(h), hence point (h,k) represents a minimum point when a is positive and a maximum point when a is negative. This point is called the vertex of the graph of f.

Example: Find the vertex of the graph of each function and identify it as a minimum or maximum point.
a) f(x) = -(x + 2)2 - 1
b) f(x) = -x2 + 2
c) f(x) = 2(x - 3)2

a) f(x) = -(x + 2)2 - 1 = -(x - (-2))2 - 1
a = -1 , h = -2 and k = -1. The vertex is at (-2,-1) and it is a maximum point since a is negative.

b) f(x) = -x2 + 2 = -(x - 0)2 + 2
a = -1 , h = 0 and k = 2. The vertex is at (0,2) and it is a maximum point since a is negative.

c) f(x) = 2(x - 3)2 = 2(x - 3))2 + 0
a = 2 , h = 3 and k = 0. The vertex is at (3,0) and it is a minimum point since a is positive.. this should help a little

Answer 10

Mathmale I agree with u ~ the x-value is the critical value but the question asks for the maximum of minimum value which is the k. The k is our maximum or minimum value. The h just determines at what point the maximum or minimum value occurs.

Answer 11

Glad to be in agreement!

Case 1: No calculus
We determine the coordinates of the vertex from the given equation, a(x-h)^2+k. In this particular case, the vertex is at (h,k); and k, whether positive or negative, is the y-component of the vertex. Should it happen that a is positive, the graph opens upward and we have a local and absolute minimum at (h,k). Should a be neg., the graph opens downward and we have a local and absolute maximum at (h,k). If all we care about is the y-component, then yes, the value of k is the y-comp. of the vertex, or, in other words, the y-comp of the min. or max.

Case 2: Using calculus to find the minimum or maximum of the function a(x-h)^2+k
Find the derivative of y=a(x-h)^2+k with respect to x, and set the resulting expression equal to zero: (dy/dx)=2a(x-h) = 0. This results in x=h, which in turn is the x-coordinate of the vertex or minimum/maximum. Were we to substitute this into the function y=a(x-h)^2+k, we'd necessarily end up with the y-value, k: y=a(h-h)^2 + k.