defining_quadratic_functions.md

Defining the quadratic function

Defining the function from Points

To define the function we can use the basic formula of quadratics and the Vertex Formula: $$ \begin{align*} f(x) &= ax^2 + bx + c \rightarrow \text{Basic Formula} \ \text{And} \ f(x) &= a(x - u)^2 + v \rightarrow \text{Vertex Formula} \end{align*} $$ There are three ways to define it:

1. This has 3 Unknown values, which means we need 3 equations | Information is needed.
2. Vertex is given, which means the vertex formula only has a single unknown variable. Which means one additional point is needed to define the functional function.
3. Is there a third way?

Example 1

Define a function, so that it goes through the points $P(1, -1)$, $Q(2, 4)$ and $R(4, 8)$

Basic Formula: $\displaystyle y = ax^2 + bx + c$ Point P: $\displaystyle -1 = a * 1^2 + b * 1 + c$ Point Q: $\displaystyle 4 = a * 2^2 + b * 2 + c$ Point R: $\displaystyle 8 = a * 4^2 + b * 4 + c$

$$ {\left| \begin{align*} -1 &= a + b + c \\ 4 &= 4a + 2b + c \\ 8 &= 16a + 4b + c \end{align*} \right| \text{Solve 3x3 System of Equations (SoE)} } $$

$$ {\left| \begin{align*} -1 &= a + b + c \\ -5 &= -3a - b \\ 4 &= 12a + 2b \end{align*} \right| \begin{align*}\\ (I - II) = (IV) \\ (III - II) = (V) \end{align*} } $$

$$ {\left| \begin{align*} -1 &= a + b + c \\ -5 &= -3a - b \\ -6 &= 6a \end{align*} \right| \begin{align*} \\ \\ (IV * 2)+(V) \\ \end{align*} } $$

$$ \begin{align*} a &= \underline{-1} \\ b &= -3a + 5 = 3 + 5 = \underline{8} \\ c &= -1 - a - b = -1 + 1 - 8 = \underline{-8} \\ \rightarrow f(x) &= \underline{\underline{-x^2 + 8x - 8}} \end{align*} $$

Example 2

Define a function, so that it goes through the points $P(-4, 8)$, $Q(0, 0)$ and $R(10, 15)$

Basic Formula: $\displaystyle y = ax^2 + bx + c$ Point P: $\displaystyle 8 = a * -4^2 + b * -4 + c$ Point Q: $\displaystyle 0 = a * 0^2 + b * 0 + c$ Point R: $\displaystyle 15 = a * 10^2 + b * 10 + c$

$$ {\left| \begin{align*} 8 &= 16a + -4b + c \\ 0 &= c \\ 15 &= 100a + 10b + c \end{align*} \right| \text{Solve 3x3 System of Equations (SoE)} } $$

$$ {\left| \begin{align*} 8 &= 16a + -4b \\ 15 &= 100a + 10b \end{align*} \right| \text{Multiply I by 5 and II by 2} } $$

$$ {\left| \begin{align*} 40 &= 80a - 20b \\ 30 &= 200a + 20b \end{align*} \right| \text{Add I and II} } $$

$$ {\left| \begin{align*} 70 = 280a \end{align*} \right| \text{Shorten it} } $$ $$ {\left| \begin{align*} \frac{1}{4} = a \end{align*} \right| \text{Replace this, in one of the previous functions, to get B} } $$

$$ {\left| \begin{align*} 8 &= 16a + -4b \\ &\downarrow \\ 8 &= 16 * \frac{1}{4} - 4b \\ 8 &= 4 - 4b \\ 4 &= -4b \\ 4b &= -4 \\ b &= -1 \end{align*} \right| \text{Multiply I by 5 and II by 2} } $$

Defining function from 2 points

Example 1

Define a function, so that it goes through the points $S(-1, 0)$ and $P(1, 3)$

Vertex formula: $f(x) = a(x - u)^2 + v$

Fill in values from points: $f(x) = a(x + 1)^2 + 0 = a(x + 1)^2$

Fill in $P$: $3 = a(1 + 1)^2 = a * 2^2 = 4a$ Calculate $a = \frac{3}{4}$ Then: $f(x) = \frac{3}{4}(x + 1)^2$ And then convert to basic formula:

$$ \begin{align*} f(x) &= \frac{3}{4}(x^2 + 2x + 1) \\ &= \underline{\underline{\frac{3}{4}x^2 + \frac{3}{2}x + \frac{3}{4}}} \end{align*} $$

Example 2

Define a function, so that it goes through the points $S(2, 1)$ and $P(0, 4)$

Vertex formula: $f(x) = a(x - u)^2 + v$

Fill in values from points: $f(x) = a(x - 2)^2 + 1 = a(x - 2)^2 + 1$

Fill in $P$: $4 = a(0 - 2)^2 + 1 = a * 4 + 1$ Calculate $a = \frac{4 - 1}{4} = \frac{3}{4}$ Then: $f(x) = \frac{3}{4}(x - 2)^2 + 1$ And then convert to basic formula:

$$ \begin{align*} f(x) &= \frac{3}{4}(x^2 - 4x + 4) + 1 \\ &= \frac{3}{4}x^2 - 3x + 3 + 1 \\ &= \underline{\underline{\frac{3}{4}x^2 - 3x + 4 }} \end{align*} $$