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Lorico [155]
2 years ago
7

Graphing f(x)=(x+2)^2-3 in vertex form

Mathematics
1 answer:
USPshnik [31]2 years ago
5 0

Answer:

Step-by-step explanation:

This is a parabola shaped like a U.

The minimum value is at (-2, -3).

Find the zeroes:

(x + 2)^2 = 3

Find the zeroes:

x + 2 = +/- √3

x = +/-√3 - 2

x = -0.27, -3.73.

So the graph cuts the x axis at (-0.27, 0)  and (-3.73, 0)

when x = -4 , f(x) = 1 and when x = 1,  f(x) =  6.

So you can now draw the curve through these 5 points and it will be shaped like a U, symetrical about the line x = -2

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The given parameters can be represented as:

\begin{array}{ccccccc}x & {5} & {4} & {3} & {2} & {1}& {0} & P(x) & {25\%} & {30\%} & {20\%} & {15\%} & {5\%} & {5\%} \ \end{array}

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