Question

How do you tell whether the graph opens up or down, find the vertex, and find the axis of symmetry given `y=1/10x^2+x-2`?

Answer 1

The graph 'opens upwards'.

Vertex`->(x,y)=(-5,-9/2)`

Axis of symmetry `->x=-5`

Explanation:

`color(blue)("Opens up or down")`

If the coefficient of `x^2` (the number in front of it :could be 1 )
is positive then the graph is of form `uu`

If the coefficient of `x^2` is negative then the graph is of form `nn`

In this case the coefficient is `+1/10` so it is of form `uu->` opens up.
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`color(blue)("Determine the vertex")`

There are several ways but I am going to show you a shortcut. The method forms part of the process for 'completing the square.

Consider the standardised form of `y=ax^2+bx+c`

Write as `y=a(x^2+b/ax)+c`

Note that `axxb/a=b`

`x_("vertex")=(-1/2)xxb/a`

We have the equation:
`y=1/10x^2+x-2`

Where `a=1/10 and color(red)(b=1) -> "from "x=color(red)(1)x`

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

`x_("vertex")=-5`

Substitute for `x=-5` to find `y_("vertex")`

`y_("vertex")=1/10(-5)^2+(-5)-2`

`y_("vertex")=2.5-5-2 = -4.5 ->-9/2`

Vertex`->(x,y)=(-5,-9/2)`
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`color(blue)("Determine axis of symmetry")`

This is a line that is parallel to the y-axis and passes through `x_("vertex")`

So axis of symmetry is `x=-5`