# How do you tell whether the graph opens up or down, find the vertex and the axis of symmetry of y=(x-7)(x-1)?

##### 1 Answer
Apr 16, 2017

Vertex$\to \left(x , y\right) = \left(4 , - 9\right)$
Axis of symmetry $\to x = 4$

#### Explanation:

$\textcolor{b l u e}{\text{General shape of the graph}}$

Consider the standardised quadratic form of $y = a {x}^{2} + b x + c$

If $a$ is positive the general shape of the graph is $\cup$
If $a$ is negative the general shape of the graph is $\cap$

Lets multiply out the brackets:

$y = \textcolor{b l u e}{\left(x - 7\right)} \textcolor{g r e e n}{\left(x - 1\right)}$

Multiply everything in the right bracket by everything in the left.

color(green)(y=color(blue)(x)(x-1)color(blue)(" "-7)(x-1)

$y = {x}^{2} - x \text{ } - 7 x + 7$
$\textcolor{w h i t e}{.}$

$y = {x}^{2} - 8 x + 7$

So for $y = a {x}^{2} + b x + c$

$a = 1 \text{; "b=-8"; } c = 7$

So $a = + 1$ which is positive so the graph is of type: $\cup$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{To determine the vertex - a sort of cheat way}}$

Consider the form $y = {x}^{2} - 8 x + 7$

Write in the form of $y = a \left({x}^{2} + \frac{b}{a} x\right) + c$

Where ${x}_{\text{vertex}} = \left(- \frac{1}{2}\right) \times \frac{b}{a}$

=>color(green)(x_("vertex")="axis of symmetry" = (-1/2)xx(-8)=+4)

Substituting $x = 4$

y=x^2-8x+7" "->" "y_("vertex")=(4)^2-8(4)+7

${y}_{\text{vertex}} = 16 - 32 + 7$

$\textcolor{g r e e n}{{y}_{\text{vertex}} = - 9}$

color(blue)("Veretex"->(x,y)=(4,-9)