Getting the quadratic function given the vertex and one point.

Question

Find out the quadratic function for the parable that contains the point $(1,1)$ and the vertex $(-2,3)$.

The notes I got are pretty vague:

$$b = 4\frac{-2}{9} = \frac{-8}{9}\\ c = -\frac{19}{9}\\ a = -\frac{2}{9}$$

So

$$f(x) = -\frac{2}{9}x^2-\frac{8}{9}+\frac{19}{9}$$

Didn't really help me. Why did they multiply $4$ by $\frac{-2}{9}$ to get $b$? And of course, how were $c$ and $a$ calculated?

Answer 1

First let the quadratic function be represented generically by $f(x) = ax^2 + bx + c$

You know that $(1,1)$ and $(-2,3)$ are points on this curve, letting you write the equations:

$a(1)^2 + b(1) + c = 1 \implies a + b + c = 1 $ [equation 1]

$(-2)^2a + (-2)b + c = 3 \implies 4a -2b + c = 3$ [equation 2]

Finally, you know that a quadratic reaches either a minimum or a maximum at its vertex, i.e. the vertex is a stationary point, and that means the derivative evaluated at the vertex is zero.

$f'(x) = 2ax + b$

Set $f'(-2) = 0$ to give:

$2a(-2) + b = 0 \implies -4a + b = 0 \implies b = 4a$ [equation 3]

Solve those 3 equations simultaneously to see what you get.

What your "vague" notes gave is really indeed too vague to make much sense of without working the problem out. But it's clear that the first line is using the result from my equation 3, for instance.

---

If you wish to eschew calculus, you can also do it by using this notion: the quadratic parabola is symmetrical about a vertical line through the vertex. The horizontal distance (i.e. $x$ units) between $(-2,3)$ and $(1,1)$ is $3$. Which means that the point $3$ units to the left of $(-2,3)$ (and with the same $y$-coordinate as $(1,1)$) will also lie on the curve. That point is $(-5,1)$.

You can now use $(-5,1)$ to form the third equation to give you the same solution.

Answer 2

If a quadratic function has vertex $(h,k)$, then it can be written as $$ f(x) = a (x - h)^2 + k $$ where $a$ is some other real number. (This is one of the things they usually expect you to just memorize.) So since the vertex is $(-2, 3)$, you get $$ f(x) = a ( x - (-2))^2 + (3) = a ( x + 2)^2 + 3. $$ Now, we use $f(1) = 1$ to solve for $a$. Plugging in $x = 1$, we get \begin{align*} (1) &= a((1) + 2)^2 + 3 \\ 1 &= 9a + 3 \\ -2 &= 9a \\ a &= -\frac{2}{9} \end{align*}

Therefore, we have $$ f(x) = - \frac{2}{9} (x + 2)^2 + 3. $$ We can also expand this out to get the usual form $ax^2 + bx + c$: \begin{align*} - \frac{2}{9} (x + 2)^2 + 3 &= -\frac{2}{9} (x^2 + 4x + 4) + 3 \\ &= -\frac{2}{9} x^2 - \frac{8}{9} x - \frac{8}{9} + 3 \\ &= -\frac{2}{9} x^2 - \frac{8}{9} x - \frac{8}{9} + \frac{27}{9} \\ &= \boxed{-\frac{2}{9} x^2 - \frac{8}{9} x + \frac{19}{9}}. \end{align*}