Question

How do you find the y intercept, axis of symmetry and the vertex to graph the function `f(x)=-0.25x^2+5x`?

Answer 1

Vertex`->(x,y)=(10,25)`
`y_("intercept")->(x,y)=(0,0)`
`x_("intercept") = (x,y)=(0,0) and (20,0)`
Axis of symmetry `->x_("vertex")=10`

Explanation:

As the coefficient of `x^2` is negative the graph is of format `nn`
so we have a maximum.

`color(blue)("Determine the vertex")`

Set as `y=-0.25x^2+5x`

Write as:`" "y=-0.25(x^2-20x)`

`x_("vertex")=(-1/2)xx-20 = +10`

By substitution:`" "y_("vertex")=-0.25(10)^2+5(10) = -25+50=+25`

Vertex`->(x,y)=(10,25)`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Determine the y-intercept")`

Compare to the standard form `y=ax^2+bx+c`

In the given function there is no value given for `c` so we must have: `c=0`.

So: `y_("intercept")->(x,y)=(0,0)`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Determine the x-intercept")`

If `y_("intercept")=(0,0)` then this also equals one of the x-intercepts.

Set `y=0=-0.25x^2+5x`

`=>0=x(-0.25x+5)`

The other x-intercept is at:

`0.25x=5" "->" "x=5/0.25=20`

`x_("intercept") = 0 and 20`