Question

How do find the vertex and axis of symmetry, and intercepts for a quadratic equation `y=x^2+3x-5`?

Answer 1

`"axis of symmetry is "x=-3/2`
`x_("intercept")~~-4.19 and x_("intercept")~~1.19` to 2 decimal places
`y_("intercept")= -5`

Explanation:

You have several approaches available.

`color(blue)("Axis of symmetry - a sort of cheat way")`

Let me show you a quick way of determining the axis of symmetry.

Standardised form of `y=ax^2+bx+c`

Write as

`y=a(x^2+b/ax) +c color(white)("ddd") ->color(white)("ddd") y=1(x^2+3/1x)-5`
This is the beginnings if completing the square

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

`color(magenta)(x_("vertex")="axis of symmetry" = (-1/2)xx3=-3/2)`

Another way is the determine the x-intercepts (if there is any) the axis of symmetry is 1/2 way between those 2 points.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Determine the x-intercepts")`

`y=x^2+3x-5`

`x=(-b+-sqrt(b^2-4ac))/(2a) -> x=(-3+-sqrt(3^2-4(1)(-5)))/(2(1))`

`x=-3/2+-sqrt(29)/2`

`x=-4.19258.... and x=+1.19258...`

`x_1~~-4.19 and x_2~~1.19` to 2 decimal places

Just to check the axis of symmetry

`color(magenta)((x_1+x_2)/2= -1.5=-3/2 larr" confirmed")`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Determine the y-intercepts")`

This occures at `x=0`

`y_("intercept")=0^2+3(0)-5 = -5`