What is the axis of symmetry and vertex for the graph #y=4x^2+5x-1#?

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Answer 1 · 2017-07-05T13:54:01

#x_("vertex")=" axis of symmetry"=-5/8#

Vertex#->(x,y)=(-5/8,-41/16)#

The coefficient of #x^2# is positive so the graph is of form #uu#. Thus the vertex is a minimum.

#y=4x^2+5x-1" "...........................Equation(1)#

#color(green)(ul("Part"))# of the process of completing the square gives you:

#y=4(x^2+5/4x)-1" "....................Equation(2)#

#x_("vertex")=(-1/2)xx(+5/4)=-5/8#

Substitute for #x" in "Equation(1)# giving:

#y_("vertex")=4(-5/8)^2+5(-5/8)-1#

#y_("vertex")=-2 9/16 ->-41/16#

Vertex#->(x,y)=(-5/8,-41/16)#