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Assume that the function given by $ f(x) = ax^2 + bx + c $, $ a \neq 0 $

has two real zeros. Show that the $ x $ -coordinate of thevertex of the graph is the average of the zeros of $ f $.(Hint: Use the Quadratic Formula.)

This problem is the special case for which$$f\left(x_{1}\right)=f\left(x_{2}\right)=0$$Visually it is easy to understand that as $x=-\frac{b}{2 a}$ is an axis of symmetry then the midpoint of two points that have the same value lies on that axis.

Algebra

Chapter 2

Polynomial and Rational Functions

Section 1

Quadratic Functions and Models

Quadratic Functions

Complex Numbers

Polynomials

Rational Functions

Missouri State University

Campbell University

Harvey Mudd College

Lectures

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In mathematics, the absoluâ€¦

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Proof Assume that the funcâ€¦

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(a) determine the real zerâ€¦

03:36

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03:34

for this problem were given a second degree polynomial There has to really zeros. So this is key here. This means that if you look at the discriminative, this is B squared minus four a. C The thing that goes into in the radical for the quadratic formula this has to be bigger than zero. And so that means that the roots to to really zeros or given by the following the first one. Let's call it X one So we'LL go to the contractor formula Negative B This one will do the plus B squared minus four a c all over two a and then x two This will be the other roof from the quadratic This time we'LL use the minus sign on the radical So these are two rials zeros and that you could see their distinct Here you're adding a positive number The radicals positive because of this Where is an ex to your subtracting that positive number And now we want to show that the x coordinate of the Vertex Well, the Vertex is always given by negative B over two a comma f of negative B over two a and so that first term here in the parentheses. This's the X coordinate of the Vertex, please. And so the question here is whether this thing is the average of X one and X two show that the X coordinates of the Vertex That's this is the average of X one and X two. So we'LL go ahead and take the average of X one and next to so we'LL add them together Divide by two So now we're adding these two radical or these two expressions over here And when you add those two together you could go ahead and combined the denominator. So first, let me just write out the expressions and then I'll divide by two at the end So this X one plus X who divided by two Now they both have the same denominator two way so we can go ahead and combine these numerator tze And when we do those radicals have different signs so they cancel and you get negative two b over two a and we still have our one half and simplifying That gives you negative B over two a and this is as we mentioned before the ex corner of the Vertex. So this completes the problem

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