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In Exercises 17-34, sketch the graph of the quadratic function without using a graphing utility. Identify the vertex, axis of symmetry, and x-intercept(s).

$ h(x) = 12 - x^2 $

Vertex: $\quad(0,12)$Axis of symmetry: $\quad x=0$$x$ -intercepts: $\quad( \pm 2 \sqrt{2}, 0)$

Algebra

Chapter 2

Polynomial and Rational Functions

Section 1

Quadratic Functions and Models

Quadratic Functions

Complex Numbers

Polynomials

Rational Functions

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Let's graph this quadratic up twelve minus x square. So here there's a few ways telegraph office. One of them is You can just take the take the graph of X squared and then multiply that by a negative one that will reflect the graph along the X axis. And then the last step would be to just shift that whole graph up by twelve units and so it will go ahead and do that over here. So if you plug in X equals zero, we get out twelve. So maybe go by twos. Here six a. Ten twelve If you plug in one or native one, you'LL get eleven. So here we're going by one's on the X axis. So it's plugging a few more points here. He plugged in to you'LL get out How about if you plug in three You okay each of three twelve minus three square and that will give you three as well. And then how about when you plug in for when you plug it in for it's for minus sixteen? And so we could see that we'll get some negative values here on the graph Negative fours over here and negative for here is well, And then let's just give a really rough, rough graph here. Should be passing through those points. But it wouldn't really look like a problem. So let me just exaggerate here a little bit so we can see that we have intercepts between negative for negative three and between three and four. Well, have we can actually find out what those are in a moment. But here we did the first part. We discussed the graph without using a cracking utility. Next up find the vortex. Well, in this case, you can just look at the graph and see that the point here is zero twelve. But in general, you can always use the following where the and the beer coming from the coefficients of your quadratic. But in our case, we see that there's no ex term. So here we just have a is negative one be a zero and we don't even need to see so I won't mention that, but it's well there. But as you can see here on the Vertex is you don't you don't need to see And if you plug in the be in the way here you'll end up getting the point zero and then h zero as well. So that takes care of the Vertex. Access is imagery that is always given by the equation. X equals negative B over two A. So, in our case, that's just becomes X equals zero negative zero over negative too, which is your zero. So that takes care of this part over here. And then we have one last part here the ex intercepts. We have to actually find the exact values of these roots. So to do that, we just said H of X, which is twelve minus x square, equal to zero, and then saw for X. So then don't forget the plus or minus here, square room. And then you could pull right. That is four times three and then take out that four outside the radical. So that gives us the intercepts. Ex intercepts. Negative. Two, three and zero. That's this one over here, the negative one. Then we have positive tune three and zero. That one's over here. The positive one. So we have the graph, the Vertex accesses, symmetry, and then X intercepts down here. Thanks. Intercepts these points. So that's our final answer.

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