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choli [55]
1 year ago
6

2.1 & 2.2 Assessment: Vertex & Standard Form of Quadratic Functions

Mathematics
1 answer:
Brums [2.3K]1 year ago
6 0

~~~~~~\textit{vertical parabola vertex form} \\\\ y=a(x- h)^2+ k\qquad \begin{cases} \stackrel{vertex}{(h,k)}\\\\ \stackrel{"a"~is~negative}{op ens~\cap}\qquad \stackrel{"a"~is~positive}{op ens~\cup} \end{cases} \\\\[-0.35em] ~\dotfill

\begin{cases} h=1\\ k=-9 \end{cases}\implies y=a(x-1)^2 - 9\qquad \textit{we also know that} \begin{cases} x=2\\ y=-7 \end{cases} \\\\\\ -7=a(2-1)^2-9\implies 2=a(1)^2\implies 2=a \\\\[-0.35em] ~\dotfill\\\\ ~\hfill {\Large \begin{array}{llll} y=2(x-1)^2 -9 \end{array}} ~\hfill

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Given:

The set of ordered pairs in the options:

(a) (2, 3), (2, 2), (2, 4), (2, 9), (2, –2)

(b) (1, 5), (2, 6), (1, 6), (4, 5), (1, 4)

(c) (1, 0), (2, 0), (1, 3), (4, 3), (5, 7)

(d) (5, 1), (7, 3), (9, 5), (11, 7), (13, 9)

To find:

The set of ordered pairs that could represent a function.

Solution:

A set of ordered pairs represent a function, if there exist unique output (y-value) for each input (x-value).

In option (a) we have, five output values y=3,2,4,9,-2 for single input x=2. So, it is not a function.

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In option (c) we have, two output values y=0,3 for single input x=1. So, it is not a function.

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<u>We want app prices (x's) when profit (P(x)) is 0, so plugging in into the equation:</u>

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