Question

If y = x^2 - 3x + 1 , find the minimum value of y.

Answer 1

Answer:

The most straight forward way, without calculus, of finding the minimum or maximum value of a quadratic is to express it in vertex format, by completing the square on the standard form.

y = x^2 + 3x + 1

Because you have x^2 instead of ax^2, we can use an abbreviated method.

Take half of the x-term coefficient and square it. Add that number after the x-term and then subtract it, so we don't actually change the value of the expression.

y = x^2 + 3x + (3/2)^2 - (3/2)^2 + 1

y = x^2 + 3x + 9/4 - 9/4 + 1

The first three terms form a perfect square. To get that perfect square, use x and half of the coefficient of the x-term. You might as well simplify the other two terms outside the square.

y = (x + 3/2)^2 - 5/4

The minimum value is -5/4 at (value inside the square = 0) x = -3/2.

For questions where the coefficient of x^2 is not 1, use the full method. For example ( a little different from yours, but shows what to do with the constant term):

y = 3x^2 + 2x - 5

Factor the x^2 coefficient out of the first two terms.

y = 3(x^2 + 2/3x) - 5

Take half the coefficient of x, square it and add and subtract the square inside the brackets.

y = 3(x^2 + 2/3x + (1/3)^2 - (1/3)^2) - 5

y = 3(x^2 + 2/3x + 1/9 - 1/9) - 5

The first three terms are the perfect square. That -1/9 at the end is in the way, so move it out of the brackets. Remember to multiply by the constant out front as we move it out.

y = 3(x^2 + 2/3x + 1/9) - 1/3 - 5

Now what is in the brackets is the perfect square as before and this is also a good time to simplify the constants at the end.

y = 3(x - 1/3)^2 - 16/3

The minimum is -16/3 at x = 1/3.

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