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

1 Answer
Mar 7, 2018

#"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#