You can find the #"zeroes (x-intercepts), y-intercepts, and the vertex."#
The standard form of a quadratic equation is
#ax^2+bx+c#
with #c# being the #"y-intercept"#.
The #"y-intercept"# of this equation is #-7# as #c# is #-7#.
To find the #"zeroes"#, plug #0# for #y#. Use the quadratic formula.
#(-b+-sqrt(b^2-4ac))/(2a)#
Identify #a,b,"and"# #c#:
#a=2#
#b=-13#
#c=-7#
Plug in:
#(-(-13)+-sqrt((-13)^2-4*2*-7))/(2*2)=>#
#(13+-sqrt(225))/4=>#
#(13+-15)/4#
The zeroes are
#7,-1/2#
This was also factorable, as a discriminant (#b^2-4ac#) that is a perfect square tells us that the equation is factorable (#225# is a perfect square).
You could factor and get
#(2x+1)(x-7)=>#
#x=7,-1/2#
To get the vertex of an equation #ax^2+bx+c#, use:
#h=(-b)/(2a)#
#k=c-(b^2)/4a#
with #(h,k)# being the vertex:
#a=2#
#b=-13#
#c=-7#
#h=(-(-13))/(2*2)=>13/4#
#k=-7-((-13)^2)/(4*2)=>-56/8-169/8=>-225/8#
The vertex is #(13/4,-225/8)#
#----------------#
Summary:
#"y-intercept":(0,7)#
#"x-intercepts": (7,0) and# #(-1/2,0)#
#"vertex":(13/4,-225/8)#
