Which equation is y = 3(x - 2)² - (x - 5)² rewritten in vertex form? 
y = 3(x minus seven-halves) squared minus Start Fraction 27 Over 4 End Fraction
y = 2(x minus 1) squared minus 11
y = 2(x minus one-half) squared minus Start Fraction 53 Over 4 End Fraction
y = 2(x minus one-half) squared minus Start Fraction 27 Over 2 End Fraction

Solution:

The vertex form of the equation of a parabola is

f(x) = a (x - h)² + k

Where (h, k) is the vertex of the parabola

It is given that

y = 3(x - 2)² - (x - 5)²

Using the algebraic identity (a - b)² = a² - 2ab + b²

y = 3 (x² - 4x + 4) - (x² - 10x + 25)

y = 3x² - 12x + 12 - x² + 10x - 25

By further calculation

y = 3x² - x² - 12x + 10x + 12 - 25

y = 2x² - 2x - 13

Taking out 2 as common

y = 2(x² - x - 13/2)

Let us add and subtract 1/2

y = 2(x² - x + 1/4 - 1/4 - 13/2)

y = 2 [(x - 1/2)² - (1 + 26)/4]

So we get,

y = 2 [(x - 1/2)² - 27/4]

y = 2 [(x - 1/2)²] - 27/2

Therefore, the vertex form is y = 2 (x minus one-half) squared minus Start Fraction 27 Over 2 End Fraction.

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Which equation is y = 3(x - 2)² - (x - 5)² rewritten in vertex form?

Summary:

The equation y = 3(x - 2)² - (x - 5)² rewritten in vertex form is y = 2 (x minus one-half) squared minus Start Fraction 27 Over 2 End Fraction.