Question

The standard form of a quadratic equation is given as y=ax^2+bx+c and vertex form is given as y=a(x−h)^2+k. Write a formula that relates the values of a,b, and c in y=ax^2+bx+c to the values of a,h, and k in y=a(x−h)^2+k. For instance, h=ax+b would be a formula that relates h to a and b.

Answer 1

Answers:

$h = -\frac{b}{2a}\\\\k = \frac{-b^2+4ac}{4a}$

where 'a' cannot be zero.

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Explanation:

The vertex is (h,k)

The x coordinate of the vertex is h which is found through this formula

$x = -\frac{b}{2a}$

For example, if we had the quadratic $y = 3x^2-6x+5$, then we'll plug in a = 3 and b = -6 to get: $h = -\frac{b}{2a} = -\frac{-6}{2*3} = 1$

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To find the value of k, we plug that h value into the original standard form of the quadratic and simplify.

$y = ax^2+bx+c\\\\k = ah^2+bh+c\\\\k = a\left(\frac{-b}{2a}\right)^2+b\left(\frac{-b}{2a}\right)+c\\\\k = a*\frac{b^2}{4a^2}+\frac{-b^2}{2a}+c\\\\k = \frac{b^2}{4a}+\frac{-b^2}{2a}+c$

$k = \frac{b^2}{4a}+\frac{-b^2}{2a}*\frac{2}{2}+c*\frac{4a}{4a}\\\\k = \frac{b^2}{4a}+\frac{-2b^2}{4a}+\frac{4ac}{4a}\\\\k = \frac{b^2-2b^2+4ac}{4a}\\\\k = \frac{-b^2+4ac}{4a}$

It's interesting how we end up with the numerator of $-b^2+4ac$ which is similar to $b^2-4ac$ found under the square root in the quadratic formula. There are other ways to express that formula above. We need $a \ne 0$ to avoid dividing by zero. The values of b and c are allowed to be zero.