Question

How do you find the axis of symmetry, graph and find the maximum or minimum value of the function `f(x)=-3x^2+x-5`?

Answer 1

Axis of symmetry is at `x=1/6`, the maximum value at `-59/12` and minimum value at `-oo`.

Explanation:

The axis of symmetry is basically the `x`-coordinate of the vertex. The `x`- coordinate of a parabola is given by:

`-b/(2a)`, and here:

`a=-3` and `b=1`. Inputting:

`A_s=-b/(2a)`

`A_s=-1/(2*-3)`

`A_s=-1/-6`

`A_s=1/6`

Since the leading coefficient of the equation is negative, the graph opens downward. Therefore, the maximum value is the `y` coordinate of the vertex. We have:

`y=-3x^2+x-5`, and the vertex has `x=1/6`, so:

`y=-3(1/6)^2+1/6-5`

`y=-3(1/36)+1/6-5`

`y=-1/12+1/6-5`

`y=-59/12`

The minimum value is so `-oo`, as the parabola keeps going down.

Graphing such, we can confirm:

graph{-3x^2+x-5 [-9.63, 10.37, -10.64, -0.64]}

Answer 2

General shape `nn` thus a maximum

`y_("intercept")=-5`

`"No "x_("intercepts")`

Vertex `->(x,y)=(1/6,-59/12)`

Explanation:

Set `y=f(x)=-3x^2+x-5`

Write as `color(green)(y=-3(x^2color(red)(-1/3)x)-5)`
The above is the start of completing the square

`color(white)("d")`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(magenta)("The axis of symmetry is at")`
`x_("vertex")=(-1/2)xx(color(red)(-1/3)) = color(magenta)(+1/6)`

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(magenta)(y" intercept")`
Consider the form `y=ax^2+bx+c`

`y_("intercept")=c=-5`

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(magenta)("General shape of the graph")`

As the `x^2` term is negative the general shape is `nn`

Thus the vertex is a maximum.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(magenta)("Determine the vertex")`

Found above that `x_("vertex")=1/6`

`color(green)(y=-3color(red)(x)^2+color(red)(x)-5 color(white)("ddd")->color(white)("ddd")y_("vertex")=-3(color(red)(1/6) )^2+color(red)(1/6)-5`

`color(green)(color(white)("dddddddddddddddddd")->color(white)("dd")y_("vertex") =-59/12 )`

As the graph is of form `nn and y_("vertex") =-59/12` then the vertex is below the x-axis so the is NO x-intercepts.

Vertex `->(x,y)=(1/6,-59/12)`

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`color(blue)("Foot note")`

They would be expecting you to solve for the vertex by completing the square.