# Standard form to vertex form??

Mar 29, 2018

Complete the square

#### Explanation:

We want to go from y intercept form $f \left(x\right) = a {x}^{2} + b x + c$ into vertex form $f \left(x\right) = a {\left(x - b\right)}^{2} + c$

So take the example of
$f \left(x\right) = 3 {x}^{2} + 5 x + 2$

We need to factorise the co-efficient out from the ${x}^{2}$ and separate the $a {x}^{2} + b x$ from the $c$ so you can act upon them separately

$f \left(x\right) = 3 \left({x}^{2} + \frac{5}{3} x\right) + 2$

We want to follow this rule
${a}^{2} + 2 a b + {b}^{2} = {\left(a + b\right)}^{2}$
or
${a}^{2} - 2 a b + {b}^{2} = {\left(a - b\right)}^{2}$

We know that the ${a}^{2} = {x}^{2}$ and
$2 a b = \frac{5}{3} x$ so $2 b = \frac{5}{3}$

So we just need ${b}^{2}$ and then we can collapse it down to ${\left(a + b\right)}^{2}$

so $2 b = \frac{5}{3}$ so $b = \frac{5}{6}$ so ${b}^{2} = {\left(\frac{5}{6}\right)}^{2}$

Now we can add the ${b}^{2}$ term into the equation remembering that the net sum of any additions to any equation/expression must be zero)

$f \left(x\right) = 3 \left({x}^{2} + \frac{5}{3} x + {\left(\frac{5}{6}\right)}^{2}\right) + 2 - 3 {\left(\frac{5}{6}\right)}^{2}$

Now we want to make the ${a}^{2} + 2 a b + {b}^{2}$ into ${\left(a + b\right)}^{2}$ so follow the same process as above

$f \left(x\right) = 3 {\left(x + \frac{5}{6}\right)}^{2} + \frac{72}{36} - 3 \left(\frac{25}{36}\right)$

Simply the equation
$f \left(x\right) = 3 {\left(x + \frac{5}{6}\right)}^{2} - \frac{3}{36}$

Now we have the result in standard form

Mar 29, 2018

General vertex form of a quadratic function:
$f \left(x\right) = a {\left(x + \frac{b}{2 a}\right)}^{2} + f \left(- \frac{b}{2 a}\right)$
In this formula,
$\left(- \frac{b}{2 a}\right)$ is the x-coordinate of the vertex
$f \left(- \frac{b}{2 a}\right)$ is the y-coordinate of the vertex.
To proceed, first find $x = - \frac{b}{2 a}$.
Next, find $f \left(- \frac{b}{2 a}\right)$
Example: Transform to vertex form -->
$f \left(x\right) = {x}^{2} + 2 x - 15$
x-coordinate of vertex:
$x = - \frac{b}{2 a} = - \frac{2}{2} = - 1$
y-coordinate of vertex:
$f \left(- \frac{b}{2 a}\right) = f \left(- 1\right) = 1 - 2 - 15 = - 16$
Vertex form:
$f \left(x\right) = {\left(x + 1\right)}^{2} - 16$