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A transformation T:R2→R2𝑇:𝑅2→𝑅2 has ruleT(x,y)=(3x−4,−y−12)𝑇(𝑥,𝑦)=(3𝑥−4,−𝑦−12)Find the image of the curve with equation y=2x−4−−−−√+3𝑦=2𝑥−4+3 under this transformation.

Question

A transformation T:R2→R2𝑇:𝑅2→𝑅2 has ruleT(x,y)=(3x−4,−y−12)𝑇(𝑥,𝑦)=(3𝑥−4,−𝑦−12)Find the image of the curve with equation y=2x−4−−−−√+3𝑦=2𝑥−4+3 under this transformation.

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Solution

The transformation T is given by T(x, y) = (3x - 4, -y - 12).

The curve is given by y = sqrt(2x - 4) + 3.

To find the image of the curve under the transformation, we substitute y in the transformation with the equation of the curve.

So, T(x, y) becomes T(x, sqrt(2x - 4) + 3) = (3x - 4, -(sqrt(2x - 4) + 3) - 12).

Simplify this to get the image of the curve under the transformation:

T(x, sqrt(2x - 4) + 3) = (3x - 4, -sqrt(2x - 4) - 15).

So, the image of the curve under the transformation T is the curve with equation y = -sqrt(2x - 4) - 15.

This problem has been solved

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