# How do you write a quadratic equation with Vertex- (5,5) Point- (6,6)?

Apr 15, 2016

$y = {x}^{2} - 10 x + 30$

#### Explanation:

Two solve the question the two forms of a parabola's equation should be known:
Standard form: $y = a {x}^{2} + b x + c$
Vertex form $y = a {\left(x - h\right)}^{2} + k$ (with vertex at ($h , k$))

First we input values from the question into the vertex form equation:
$y = a {\left(x - h\right)}^{2} + k$
$= a {\left(x - 5\right)}^{2} + 5$
$= a {\left(x - 5\right)}^{2} + 5$

Still there is the unknown $a$ for which we need to find its value. Using the other piece of information, we can find $a$:
$6 = a {\left(6 - 5\right)}^{2} + 5$
$6 = a + 5$
$a = 1$

Now that we know $a$, we can write an equation with no unknowns and convert it to its standard form:
$y = {\left(x - 5\right)}^{2} + 5$
$y = \left({x}^{2} - 10 x + 25\right) + 5$
$y = {x}^{2} - 10 x + 25 + 55$
$y = {x}^{2} - 10 x + 30$

Apr 15, 2016

Using the standard vertex form for a parabola with a vertical axis of symmetry:
$\textcolor{w h i t e}{\text{XXX}} y = m {\left(x - a\right)}^{2} + b$ with its vertex at $\left(a , b\right)$

A parabola with a vertical axis of symmetry and a vertex at $\left(5 , 5\right)$
would have the form:
$\textcolor{w h i t e}{\text{XXX}} y = m {\left(x - 5\right)}^{2} + 5$ for some constant value $m$

If $\left(x , y\right) = \left(6 , 6\right)$ is a point on the required equation, then
$\textcolor{w h i t e}{\text{XXX}} 6 = m {\left(6 - 5\right)}^{2} + 5$
$\textcolor{w h i t e}{\text{XXX}} m = 1$
and the equation would be:
$\textcolor{w h i t e}{\text{XXX}} y = 1 {\left(x - 5\right)}^{2} + 5$

This could be rearranged as
$\textcolor{w h i t e}{\text{XXX}} y = {x}^{2} - 10 x + 30$ (standard quadratic form)#

For a parabola with a horizontal axis of symmetry,
the $x$ and $y$ values could be exchanged giving
$\textcolor{w h i t e}{\text{XXX}} x = {y}^{2} - 10 y + 30$
as an alternative solution that meets the given requirements.