Parabola Reflections

Today's question is of a type we didn't have for a long time - pure variables, no numbers. It's very interesting, and although it looks impossible at first, one little step makes it very, very easy.

The parabola y = ax² + bx + c has its vertex on (h, k). The parabola is reflected about the line y = k. The new parabola is: y = dx² + ex + f. Find a + b + c + d + e + f.

Looks impossible, doesn't it? You can try doing some complex calculations showing that -b/2a = h (you should know that), then plug it back, and equate it to y. That's hard. However, parabolas have two forms: standard form and vertex form. Guess what - we have the vertex!

y = a(x - h)² + k

Now we can open the parentheses and find exactly a, b, and c:
y = ax² - 2axh + ah² + k

That means:
a = a
b = -2ah
c = ah² + k

When the line is reflected about y = k, it changes the coordinates. Since the vertex is on y = k, the parabola just switches sides. This means a = -a and b becomes the opposite too (2ah) to keep the vertex.

Notice that the vertex remains the same because the line was reflected about y = k. When you reflect point (x, n) about the line y = z, you end up with (x, 2z - n). If you reflect (h, k) about y = k, you get (h, 2k - k), which is (h, k).

Since each point turns into (x, 2k - y), the new y-intercept, f, becomes (0, 2k - (ah² + k)), which is (0, 2k - ah² - k), or (0, k - ah²).

a and d are opposites, so a + d = 0. b and e behave the same. The only addends left are c and f:
c + f = ah² + k + k - ah² = 2k

This big sum, a + b + c + d + e + f, equals to 2k.

Hope you liked it,
Nadav

nadavs