Problem

The profit $ P $ (in millions of dollars) for a r…

02:35

Problem 95 Hard Difficulty

Write the quadratic function

$ f(x) = ax^2 + bx + c $

in standard form to verify that the vertex occurs at

$ \left (-\frac{b}{2a} , f \left (-\frac{b}{2a} \right) \right) $.

Answer

$\left(-\frac{b}{2 a},-\frac{b^{2}}{4 a}+c\right)=\left(-\frac{b}{2 a}, f\left(-\frac{b}{2 a}\right)\right)$

Related Courses

Algebra

Precalculus with Limits

Chapter 2

Polynomial and Rational Functions

Section 1

Quadratic Functions and Models

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Video Transcript

recalls that the standard form of the quadratic is this expression over here where the Vertex is given by the point each complicate. So as usual, we'LL do some algebra here. So let's first rewrite this by factoring on a A from the first two terms And then now go ahead and complete the square inside the fantasy. So as usual, you go inside the front disease, you take the constant and from the X and then you square you divide that by two So that'LL be b over two a and then you square that so in our case and then we still have our plus c However, we need to subtract here. So notice we just added in this new term Technically, we didn't just add in this term and red because we're multiplying by a so we really added And so if we want to keep this equation, we should go ahead and also make up for it by subtracting p squared over for it. So now since we completed the square, we can write the expression in the parentheses X plus B over two a whole quantity square and then we have C minus B squared over for a So this is the standard form. So notice that we have X minus age. So in our case, this means H is negative. B over two A and that agrees with the answer here. The first coordinate. Now we see that our case should be this value over here. This is the value that we obtained in the standard form. So for the remainder of the problem, we should go ahead and show that these two blue quantities are equal. So this is okay and they're claiming this is the actual case. So now let's just go to the side and compute f of negative B over two A. So now go ahead and replace X with negative B over two A and the original equation and then simplify. So we'LL have b squared over for a minus. B squared over to a plus C so you can go ahead and combine those first two fractions and when you do, you get negative b squared over for a and then we still have our plus C. And this verifies that thes two quantities are equal. So indeed, the Vertex K is ost is equal to f of negative B over two a and this resolves the problem

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