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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).

$ f(x) = x^2 - x + \frac{5}{4} $

Vertex: $\left(\frac{1}{2}, 1\right)$Axis of symmetry: $x=\frac{1}{2}$$x$ -intercepts: None

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 go ahead and sketch this graph without using a calculator. So one way to go about this is to complete the square. So we see in this case, the number and friendly Xs minus one. And so you take half of that and you square that, and that's what we'LL go ahead and add and subtract here. So going close that front disease and then we still have five or four. But we haven't changed anything because the one over for that we ended, it will end up being canceled out over here. Now, the whole reason for doing this is that it lets us write the following. And that tells us the Vertex, because this is the standard form that does answer one of the questions so well. But encircle that. Sorry about this. That's ah positive on their big mistakes of plus one. So let's say it. This is one. This is a half, and then we'LL just go over one. And actually, since we know how to graft the traveler X squared, we can see that this crap is just the transformation. It goes one one half unit to the right and then because of this plus one on the outside we go one unit up so you could just go ahead and take your usual graph of X squared. And then we'LL just shift it over one half and then up by one. So let's go ahead and plant a few more points here. So we know that if we go over one unit on the quadratic, we go up by one. So here I go over one unit than up by one similarly, for the other direction. And you know that if you go over two units on the problem, then I'm X squared than your y value from the origin the Vertex goes up by four. So in this case, if we go two units over from the Vertex, we'LL go up by four will end up at the point five and similarly over here we would go to units have left and that would push us at negative three over too. And then we would go up by Fouras. Well, starting at one and we have five points in. That's about enough to go ahead and give a rough looking sketch, very rough looking sketch here. So that completes another task. We had a sketch, the graph. And just by looking at the graph, you can see that there are no ex intercepts and that could also be verified by using the quadratic formula here. So if you use the quadratic formula, you could you could see over here you have a negative inside the radical. Oh, that's not a real number. So that means that there are no solutions. And that's what we expect by looking at the graph, because we just see that there are no eggs intercepts. Uh, excuse me there when you got in a race, that and finally we have access of symmetry that's always given by X equals negative B over two to a. So, in this case that it will just be a CZ we saw already. You could just find that over here is one B is negative one. So go ahead and plug this in, and we would just get one half so X equals one half. And that's our access of symmetry. And there's our final answer

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