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)
