Oxford Entrance Exam Problem 7

Algebra Level 3

y = 2 x 2 − 4 x + 3 \LARGE y=2^{x^2-4x+3}

How can the graph of the function above be obtained from the graph of y = 2 x 2 y=2^{x^2} ?

a translation parallel to the x-axis followed by a stretch parallel to the y-axis. reflection in the y-axis followed by translation parallel to the y-axis. a translation parallel to the x-axis followed by reflection in the y-axis. a stretch parallel to the y-axis followed by a translation parallel to the y-axis. a translation parallel to the x-axis followed by a stretch parallel to the x-axis.

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2 solutions

With f ( x ) = 2 x 2 f(x) = \large 2^{x^{2}} we have that g ( x ) = 2 x 2 − 4 x + 3 = 2 ( x − 2 ) 2 − 1 = g(x) = \large 2^{x^{2} - 4x + 3} = 2^{(x - 2)^{2} - 1} = 1 2 f ( x − 2 ) . \dfrac{1}{2}f(x - 2).

Thus to obtain g ( x ) g(x) we must shift f ( x ) f(x) to the right by 2 2 units, (i.e., a translation parallel to the x x -axis), followed by a "halving" of the shifted y y values, (i.e., a stretch parallel to the y y -axis).

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Let

We need to obtain

Completing the square, ; Simplifying,

So,

When we compare this with We see that, this function has been obtained by : Translating right by 2 units - a translation parallel to the x axis Stretching vertically by 1/2 units - a stretch parallel to the y axis

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