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This example illustrates another way of finding the vertex of a parabola (and deriving the quadratic formula). Given $y=a x^{2}+b x+c$ multiply both sides of the equation by $4 a,$ obtaining,$$4 a y=4 a^{2} x^{2}+4 a b x+4 a c$$Now add $b^{2}$ both sides of the equation yielding$$b^{2}+4 a y=4 a^{2} x^{2}+4 a b x+b^{2}+4 a c$$Note the first three terms on the right hand side are a prefect square, so we may factor to obtain,$$b^{2}+4 a y=(2 a x+b)^{2}+4 a c$$Solve for $y$ and then show that it follows that the vertex occurs when $x=-b / 2 a .$ Also show that setting $y=0$ results in the quadratic formula.

$$x^{2}+y^{2}-2 x y-8 x h-8 y h=0$$

Algebra

Chapter 1

Functions and their Applications

Section 4

Quadratic Functions - Parabolas

Functions

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Lectures

01:43

In mathematics, a function is a relation between a set of inputs and a set of permissible outputs with the property that each input is related to exactly one output. An example is the function that relates each real number x to its square x^2. The output of a function f corresponding to an input x is denoted by f(x).

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So the example illustrates another way of finding the vertex of a problem. Um Given y equals x squared plus bx plus C. We can multiply both sides of the equation by four A. And then add B squared to both sides of the equation. Then we can note that the first three terms on the right hand side are perfect square. So then we obtain that and now we want to solve for why? Um So when we do this we want to subtract b squared. So we'll have And the two x. Let's be squared. Okay? And then plus four a. c. And then we're gonna subtract the B squared and we're going to divide the whole thing by four A. So that right there um would be what Y is equal to. And we see that if we let y equal zero, it results in the quadratic formula because we would end up getting solving for action, saying that the quadratic formula.

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