# How do you find the minimum and maximum value of y=(x+7)(x+3)?

Feb 21, 2017

minimum $\to \left(x , y\right) = \left(- 5 , - 4\right)$
maximum $\to y = + \infty$

#### Explanation:

If you multiply out the brackets you have the general form of:
$y = a {x}^{2} + b x + c$ in this case $a = 1$ giving just $+ {x}^{2}$

As this is positive the graph is of general shape $\cup$
Thus the vertex is a minimum.

The vertex is $\frac{1}{2}$ way between the x-intercepts.

Set:$\text{ } y = 0 = \left(x + 7\right) \left(x + 3\right)$

Thus

$x + 7 = 0 \implies x = - 7$
$x + 3 = 0 \implies x = - 3$

So ${x}_{\text{vertex}} = \frac{- 3 - 7}{2} = - 5$

Substitute $x = - 5$ to give:

${y}_{\text{vertex}} = \left(- 5 + 7\right) \left(- 5 + 3\right) = 2 \times \left(- 2\right) = - 4$

Thus the minimum $\to \left(x , y\right) = \left(- 5 , - 4\right)$

As the graph is of form $\cup$ then the maximum is $y = + \infty$