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 $(2,8)$?

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 $(2,8)$?

Here is my attempt:

Use the vertex form of a parabola $f(x) = a(x – h)^2 + k$ where point $(h, k)$ is the vertex of the parabola.

Since the vertex is on the $x-axis$, the value of the ordinate $k$ of the vertex is $0$.

Hence, we will have $f(x) = a(x – h)^2$. I tried to substitute the values of $x$ and $y$ in each of the two given points on the parabola into the vertex form to solve for the values of $h$ and $a$.

Using point $(1, 4)$, $f(1) = a(1 – h)^2 = 4$

$(h^2 – 2h + 1)a = 4$ $(eq 1)$

Using point $(2, 8)$, $f(2) = a(2 – h)^2 = 8$

$(h^2 – 4h + 4)a = 8$ $(eq 2)$

After this, I tried to subtract $(eq 2)$ from $(eq 1)$.

$(eq 1) – (eq 2) = (h^2 – 2h + 1)a – (h^2 – 4h + 4)a = 4 – 8$

$(2h - 3)a = -4$

I am stuck after this solution. Any comments and suggestions will be much appreciated. Thank you!

Solution 1:

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 $h = \sqrt 2$

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

For $h = -\sqrt 2$

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