# How do you convert from vertex form to intercept form of y-4=-(x-4)^2?

Apr 2, 2015

The representation of the original function in intercept form is
$y = - \left(x - 2\right) \left(x - 6\right)$

#### Explanation:

Intercept form of a quadratic function, by definition, is a form
$y = k \left(x - \alpha\right) \left(x - \beta\right)$
It's called intercept form because $\alpha$ and $\beta$ are values of $x$ where $y$ equals to zero and, therefore, values where parabola that represents a graph of this quadratic function intercepts the X-axis.

In other words, $\alpha$ and $\beta$ are solutions to an equation
$k \left(x - \alpha\right) \left(x - \beta\right) = 0$

Transform our expression into traditional functional form.
$y - 4 = - {\left(x - 4\right)}^{2}$
$y = - {x}^{2} + 8 x - 12$
Now let's find the solutions of the equation
$- {x}^{2} + 8 x - 12 = 0$
or, in a simpler representation,
${x}^{2} - 8 x + 12 = 0$
Solutions are ${x}_{1} = 2 , {x}_{2} = 6$.

Therefore, representation of the original function in intercept form is
$y = - \left(x - 2\right) \left(x - 6\right)$

The graph of this function follows (notice the points where it intercepts the X-axis are $x = 2$ and $x = 6$).
graph{-(x-4)^2+4 [-10, 10, -5, 5]}

The original form of this function $y - 4 = - {\left(x - 4\right)}^{2}$ is called vertex form because it tells the location of the vertex of the parabola that represents a graph of this quadratic function - point $\left(4 , 4\right)$.

It can be easily seen if it is written as $y = - {\left(x - 4\right)}^{2} + 4$.
In this case, the rules of graph transformation tell us that the prototype function $y = - {x}^{2}$ with vertex at point $\left(0 , 0\right)$ and "horns" directed down should be shifted by $4$ to the right to represent function $y = - {\left(x - 4\right)}^{2}$ and then by $4$ up to represent function $y = - {\left(x - 4\right)}^{2} + 4$.
These two transformations shift the vertex to point $\left(4 , 4\right)$.