Given the function f(x) = 1(x - 12)^2

Given the function $f(x) = 1(x - 12)^2 - 8$ find all of the following: The vertex of the function is Does the function have a minimum or a maximum? (Type minimum or maximum) Find the extreme value of the function.

Transcript

00:01 It sounds like what you need to do is change this problem into what some people would call vertex form.

00:08 But yes, it's the same thing as standard form.

00:11 And then you can answer the questions.

00:14 So notice i put in some spaces in here because what i need to do is take half of this number and square it.

00:20 Well, half of that number is negative 4 when i square it as positive 16, which then later i need to subtract off 16.

00:29 So why did i just do that? is because now these first three are factorable.

00:36 Factors of 16 that add to be negative 8 are negative 4 and negative 4.

00:40 But instead of writing it twice, i can write 1 squared.

00:43 And then over here, 8 minus 16 is negative 8.

00:47 So this is what they're looking for for f of x equals in standard form.

00:53 So i believe that's part a.

00:56 So then the min value, well, first of all, so when this is positive in front, let me erase that.

01:06 I don't like that.

01:07 I wrote that down.

01:10 Positive will imply that you have an upwards you, an upwards parabola.

01:20 So what that means for me is that we have a min value at negative 8, and the value is here.

01:32 Sometimes we give the x coordinate.

01:34 So you might see somebody write it out as it's at 4, negative 8 is if i were to graph that problem, 1, 2, 3, 4, 1, 3, 4, 5, 6, 7, 8, this graph wouldn't open up like that.

01:50 So yeah, the min value is the y value, though...

Question

Given the function $f(x) = 1(x - 12)^2 - 8$ find all of the following: The vertex of the function is Does the function have a minimum or a maximum? (Type minimum or maximum) Find the extreme value of the function.

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