You can find the "zeroes (x-intercepts), y-intercepts, and the vertex."zeroes (x-intercepts), y-intercepts, and the vertex.
The standard form of a quadratic equation is
ax^2+bx+cax2+bx+c
with cc being the "y-intercept"y-intercept.
The "y-intercept"y-intercept of this equation is -7−7 as cc is -7−7.
To find the "zeroes"zeroes, plug 00 for yy. Use the quadratic formula.
(-b+-sqrt(b^2-4ac))/(2a)−b±√b2−4ac2a
Identify a,b,"and"a,b,and cc:
a=2a=2
b=-13b=−13
c=-7c=−7
Plug in:
(-(-13)+-sqrt((-13)^2-4*2*-7))/(2*2)=>−(−13)±√(−13)2−4⋅2⋅−72⋅2⇒
(13+-sqrt(225))/4=>13±√2254⇒
(13+-15)/413±154
The zeroes are
7,-1/27,−12
This was also factorable, as a discriminant (b^2-4acb2−4ac) that is a perfect square tells us that the equation is factorable (225225 is a perfect square).
You could factor and get
(2x+1)(x-7)=>(2x+1)(x−7)⇒
x=7,-1/2x=7,−12
To get the vertex of an equation ax^2+bx+cax2+bx+c, use:
h=(-b)/(2a)h=−b2a
k=c-(b^2)/4ak=c−b24a
with (h,k)(h,k) being the vertex:
a=2a=2
b=-13b=−13
c=-7c=−7
h=(-(-13))/(2*2)=>13/4h=−(−13)2⋅2⇒134
k=-7-((-13)^2)/(4*2)=>-56/8-169/8=>-225/8k=−7−(−13)24⋅2⇒−568−1698⇒−2258
The vertex is (13/4,-225/8)(134,−2258)
----------------−−−−−−−−−−−−−−−−
Summary:
"y-intercept":(0,7)y-intercept:(0,7)
"x-intercepts": (7,0) andx-intercepts:(7,0)and (-1/2,0)(−12,0)
"vertex":(13/4,-225/8)vertex:(134,−2258)