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#?

1 Answer
Jul 19, 2017

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.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#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)#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#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#

Tony B