Question

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

Answer 1

`P_("maximum")->(x,y)->(3/2,-3/4)`
Axis of symmetry`" "-> x=+3/2`

Explanation:

As this is in the form of `y=ax^2+bx+c" where " a=-1` then

the axis of symmetry is at

`x= (-1/2)xx3/(-1)" where the "3/(-1)" comes from " b/a`

So axis of symmetry is at `x=+3/2`

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As the `x^2` part is negative then the general curve shape is `nn` so it is a maximum. Substitute `x=+3/2` into the equation to find the value of `y_("maximum")` . I will let you work that out!

We already know that `x_("maximum")=+3/2`

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

`color(brown)("'~~~~~~~~~~~~ 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+3x-3=0`

To emphasise the point:

In the solution equation:`" "x= (b+-sqrt(b^2-4ac))/(2a)`

The part:`" " sqrt(b^2-4ac) -> sqrt(9-(4)(-1)(-3))`

giving:`" " sqrt(b^2-4ac) -> 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.