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 $(2,8)$?

Question

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!

Answer 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 $h = -\sqrt 2$

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

Answer 2

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 $\frac{(2-h)^2}{(1-h)^2}=2$, $a= 12 \pm 8\sqrt{2}$, $h=\pm \sqrt2$. The solution pairs, if expressed in the form $(a,h)$, are $(12+8\sqrt2,\sqrt2)$ and $(12-8\sqrt2,-\sqrt2)$.

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!

Answer 3

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, $$ a·(1 + \sqrt2)^2 \ = \ 4 \ \ \Rightarrow \ \ a \ \ = \ \ \frac{4}{(1 + \sqrt2)^2} \ \ = \ \ \frac{4}{3 + 2\sqrt2 } \ \ = \ \ 4·(3 - 2\sqrt2 ) \ \ \approx \ \ 0.686 $$ and $$ a'·(1 - \sqrt2)^2 \ = \ 4 \ \ \Rightarrow \ \ a' \ \ = \ \ \frac{4}{(1 - \sqrt2)^2} \ \ = \ \ \frac{4}{3 - 2\sqrt2 } \ \ = \ \ 4·(3 + 2\sqrt2 ) \ \ \approx \ \ 23.32 \ \ . $$ The graph below presents the two parabolas.