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In Exercises 35-42, use a graphing utility to graph the quadratic function. Identify the vertex, axis of symmetry, and x-intercepts. Then check your results algebraically by writing the quadratic function in standard form.

$ g(x) = x^2 + 8x + 11 $

Vertex $(-4,-5)$axis of symmetry $x=-4$$x$ -intercepts: $(-4-\sqrt{5}, 0),(-4+\sqrt{5}, 0)$

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 start by graphing this in a graphing calculator. So here we see that the Vertex is that negative for negative five. And it looks like we have approximations for ex intercepts so we can go ahead and write those down. Delicious. Record these and let's record this on the next page. And also not the access of symmetry. Looks like the Graff X equals minus four. So let's just go to the back, beat the previous page and record this and then we'LL verify this in a moment and then the eggs intercepts. These were approximations. It looks like from the calculator. So let's just go back to the calculator and you could see those roots right over here. But again, those might be approximations since it's a calculator now, the standard form. If we could find the standard form weaken, verify all of these. So we should take our previous We take that middle term aid there. So divide that got by two to get four and then he's square that ended sixteen. So we'll add sixteen but then we make up for it by subtracting. So we have not changed our polynomial. Now here I just go ahead and group those first three terms together and then eleven, minus sixteen minus five. And now we complete the square inside the fantasies. And there's our standard form X plus four to the second power minus five. Infirm, the standard form. We see that the Vertex is the point, so that agrees with the protection. The calculator. The access of symmetry is always X equals the first quarter of the Vertex, so that agrees with our access of symmetry from the computer from the graphing calculator and offer the ex intercepts. You can either work with the standard reform or the original. It's probably it's easier to work with the standard form because in this case, if you set this zero, you don't have to use the quadratic formula. Just add the five over and then take the square root. And if we go to the calculator and plug these in, we'LL get approximately this negative six point two three six and negative one point seven six four. So those air correct approximations, and that resolves the problem

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