What is $q$ in the factorized form $a(x+p)^2 + q$ of a quadratic equation?

Question

I can't put the following equation into $a(x+p)^2 + q$ form:

$$-4x^2 - 2x + 3$$

I've gotten as far as: $-4(x + (\frac{1}{4})^2)$, but what would $q$ be? I got $\frac{1}{4}$ by the fact that $p = \frac{b}{2}.$

Answer 1

Take out the factor of $-4$ from the first 2 terms to get $$-4\left(x^2 + \frac{1}{2}x \right) + 3 $$

Now note that $$x^2 + \frac{1}{2}x = \left(x + \frac{1}{4}\right)^2 - \left(\frac{1}{4}\right)^2$$ Then put this all together to get $$-4x^2 - 2x + 3 = -4\left[\left(x + \frac{1}{4}\right)^2 - \left(\frac{1}{4}\right)^2\right] + 3$$ which simplifies to $$-4x^2 - 2x + 3 = -4\left(x + \frac{1}{4}\right)^2 + \frac{13}{4}$$ and so $a = -4$, $p = \frac{1}{4}$ and $q = \frac{13}{4}$.

This is the method of completing the square.

The value of $q$ represents the $y$ coordinate of the turning point on the curve $y= -4x^2 - 2x + 3$

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Alternative 1: you could take out the factor of $-4$ from all three terms, but this would lead to messier fractions.

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Alternative 2: you could write $$-4x^2 -2x + 3 \equiv a(x+p)^2 + q $$ and expanding the RHS gives $$-4x^2 -2x + 3 \equiv ax^2 + 2apx + (ap^2 + q)$$ Then you can compare coefficients and this will give you some equations to solve for the values of $a$, $p$ and $q$