# How do you find the axis of symmetry, graph and find the maximum or minimum value of the function y=-x^2+3x-3?

Feb 16, 2016

${P}_{\text{maximum}} \to \left(x , y\right) \to \left(\frac{3}{2} , - \frac{3}{4}\right)$
Axis of symmetry$\text{ } \to x = + \frac{3}{2}$

#### Explanation:

As this is in the form of $y = a {x}^{2} + b x + c \text{ where } a = - 1$ then

the axis of symmetry is at

$x = \left(- \frac{1}{2}\right) \times \frac{3}{- 1} \text{ where the "3/(-1)" comes from } \frac{b}{a}$

So axis of symmetry is at $x = + \frac{3}{2}$

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
As the ${x}^{2}$ part is negative then the general curve shape is $\cap$ so it is a maximum. Substitute $x = + \frac{3}{2}$ into the equation to find the value of ${y}_{\text{maximum}}$ . I will let you work that out!

We already know that ${x}_{\text{maximum}} = + \frac{3}{2}$

So the point P_("maximum")->(x,y)->(3/2,-3/4)" "

$\textcolor{b r o w n}{\text{'~~~~~~~~~~~~ As foot note! ~~~~~~~~~~~~~~~~~~~~}}$

As P is a maximum and below the x-axis then we also know that that it does not cross the x-axis. So there is no solution to
$y = - {x}^{2} + 3 x - 3 = 0$

To emphasise the point:

In the solution equation:$\text{ } x = \frac{b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

The part:$\text{ } \sqrt{{b}^{2} - 4 a c} \to \sqrt{9 - \left(4\right) \left(- 1\right) \left(- 3\right)}$

giving:$\text{ } \sqrt{{b}^{2} - 4 a c} \to \sqrt{9 - 12} = \sqrt{- 3}$

As we are dealing with the sqrt of a negative number this is a sure sign that the graph does not cross the x-axis.