What is the equation of the quadratic function whose vertex of the graph is on the $x$-axis and passes through the two points $(1,4)$ and $(2,8)$?

The problem can also be interpreted geometrically; the placement of the vertex of the parabola on the $ x-$axis reduces the amount of algebra needed. Since the vertex is at $ y = 0 , $ the point $ (2 , 8) $ is twice as far "above" the $ x-$axis than $ (1 , 4) $ is. For a parabola with equation $ y = a·(x - h)^2 , $ this means that $ x = 2 $ is $ sqrt2 $ times as far from $ x = h $ than $ x = 1 $ is.

There are then two possibilities. Either the vertex is "to the left" of both points ( $ h < 1 $ ) or the vertex is between the points ( $ 1 < h < 2 ) . $ In the former case, we have $$ sqrt2·(1 - h) = 2 - h Rightarrow h = frac{2 - sqrt2}{1 - sqrt2} = -sqrt2 . $$ If the vertex is between the points, $$ sqrt2·(h - 1) = 2 - h Rightarrow h = frac{2 + sqrt2}{1 + sqrt2} = +sqrt2 . $$

The two possible quadratic functions are thus $ a·(x + sqrt2)^2 $ and $ a'·(x - sqrt2)^2 . $ We can use either of the two given points to find these coefficients: for instance, $$ a·(1 + sqrt2)^2 = 4 Rightarrow a = frac{4}{(1 + sqrt2)^2} = frac{4}{3 + 2sqrt2 } = 4·(3 - 2sqrt2 ) approx 0.686 $$ and $$ a'·(1 - sqrt2)^2 = 4 Rightarrow a' = frac{4}{(1 - sqrt2)^2} = frac{4}{3 - 2sqrt2 } = 4·(3 + 2sqrt2 ) approx 23.32 . $$ The graph below presents the two parabolas.

Let the vertex of the quadratic function be $(h,k)$. We know that $k=0$, so it is $(h,0)$. We have $a(x-h)^2+0=a(x-h)^2=y$.

We know that $a(1-h)^2=4$, and $a(2-h)^2=8$. Now just solve this system. Noting that $frac{(2-h)^2}{(1-h)^2}=2$, $a= 12 pm 8sqrt{2}$, $h=pm sqrt2$. The solution pairs, if expressed in the form $(a,h)$, are $(12+8sqrt2,sqrt2)$ and $(12-8sqrt2,-sqrt2)$.

From here you should be able to rewrite the equation to whatever form you'd like it in with a little computation.

Hope this helped!

We divide both the equations

$$frac{1}{2} = frac{h^2-2h+1}{h^2-4h+4}$$

$$h^2-4h+4 = 2h^2 - 4h+2$$

$$implies h^2 = 2 implies h = pm sqrt2$$

For $h = sqrt 2$

$$a = frac{4}{3-2sqrt2}$$

For $h = -sqrt 2$

$$a = frac{4}{3+2sqrt2}$$