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In Exercises 47-56, write the standard form of the equation of the parabola that has the indicated vertex and whose graph passes through the given point.

Vertex: $ ( -2, 5 ) $; point: $ ( 0, 9 ) $

$y=(x+2)^{2}+5$

Algebra

Chapter 2

Polynomial and Rational Functions

Section 1

Quadratic Functions and Models

Quadratic Functions

Complex Numbers

Polynomials

Rational Functions

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the standard form of equation of a problem is of this farm ffx r A, which is a constant coefficient. And then we have X minus h square plus que where huk is the Vertex. So in our problem there telling us the Vertex here So they're giving us a tch negative two and they're also giving us Kay, let me come back here and fix that. So Kay is equal to five. That's the second cornet So plugging these in It's our formula. Over here we can update our equation. We still don't know what is. But now we have X plus two inside the parentheses square that and then plus five. Now, how do we find the value of a Well, we use the fact that our graph passes through the point zero nine. So that's telling us that fo zero is time. So let's come over here and use this fact. In our case, half of zero will be a zero plus two square plus five. So this is for a plus five little sloppy there, let me come and try to fix that. So for a plus five and now we use the fact that this is equal to nine and then solve for a So we'LL go ahead and subtract that five from both sides and then divide both sides by four. And then finally, plugging this value of a back into our F gives us our final answer. So if you want, we could just write that one there X plus two. Swear Plus five. Or you could just simplify it by not writing the one. And then we can go at encircle this part right here. So that's the right hand side. Let's ignore that middle tar, and then we'LL come over here Circle that left hand side as well. That's our standard form of the problem, and that's our final answer.

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