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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: $ ( 6, 6 ) $; point: $ \left(\frac{61}{10}, \frac{3}{2} \right) $

$y=-450(x-6)^{2}+6$

Algebra

Chapter 2

Polynomial and Rational Functions

Section 1

Quadratic Functions and Models

Quadratic Functions

Complex Numbers

Polynomials

Rational Functions

McMaster University

Baylor University

University of Michigan - Ann Arbor

Lectures

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In mathematics, the absolu…

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In Exercises 47-56, write …

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Write the standard form of…

Let's write the standard form and recall what that is. This is when you write the proble in the following form where the Vertex is given by H K. So work given the birth text here six in six. So that gives us Agent K so we could at least plug in six for B both of these values in the formula and we'LL have to leave a for right now. Now we use the fact that we have a point on the graph. So this means if you plug in sixty one over ten for X, you get three over two for why? So let's plug in sixty one over ten and we could simplify this so simplified that fraction you get one over ten. However, we're told that this also equals three over too. So let's go ahead and solve this equation over here for a so that will be three minus twelve over too. And that multiply that by hundred and that gives you negative for fifty. So finally, plug in this value and for a and that'LL give us our final answer. So this is the parable a and standard form that goes through the point sixty one over ten comma three or two and the vertex is located at six comma six

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