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Olenka [21]
3 years ago
7

Solve (x - 2)^2 - 3 = 1 graphically. That is graph y = (x - 2)^2 - 3 and y = 1 on the same set of axes and find the x-value(s) o

f any points of intersection. Then use algebraic strategies to solve the equation and verify that your graphical solutions are correct.
Mathematics
1 answer:
mafiozo [28]3 years ago
4 0

Check the attachment(s) for the graphs pertaining to the question

No. 1) Check graph [first attachment]

No. 2) y = (x - 2)^2 - 3, y = 1

           (x - 2)^2 - 3 = 1,

           (x - 2)^2 = 4

            x - 2 = √4, x - 2 = - √4

            x = 4, x = 0

Substitute back to determine the respectively y-value:

y = (4 - 2)^2 - 3 = (2)^2 - 3 = 4 - 3 = 1

Check: y = (0 - 2)^2 - 3 = (-2)^2 - 3 = 4 - 3 = 1

So the points of intersection are (4, 1) and (0, 1). According to the graph, that is correct.

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Answer:

Steps in order will be

1. y=\sqrt{7x-21}\\2. y^2=7x-21 \\ 3. y^2=7(x-3)\\4. \frac{y^2}{7}=x-3\\ 5. x=\frac{y^2}{7}+3\\6.  f^{-1}(x)=\frac{1}{7}x^2+3 \:where\: x\geq 0

Step-by-step explanation:

Consider the function f(x)=\sqrt{7x-21}

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Step 2: Taking square on left side

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Step 3 : take 7 common

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Step 4 : Divide both sides by 7

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Step 5: Add 3 on both sides

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Step 6: Replace x with y and x with f^{-1}(x)

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1. y=\sqrt{7x-21}\\2. y^2=7x-21 \\ 3. y^2=7(x-3)\\4. \frac{y^2}{7}=x-3\\ 5. x=\frac{y^2}{7}+3\\6.  f^{-1}(x)=\frac{1}{7}x^2+3 \:where\: x\geq 0

Assuming one or 2 steps are missing in the diagram given.

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