Answer:
Graph I
Step-by-step explanation:
Given are two quadratic equations as
[tex]y = x^2 + 3x + 4:
y = x^2 + 2x + 1[/tex]
WE are to find the point of intersection to solve this
Since leading coefficients are positive, both parabolas will be open up.
Hence the solutions cannotbe C or D
ONly out of A and B the solutions will be there
Let us find the vertex
I parabola:
[tex]y=(x+3/2)^2+7/4[/tex]
Vertex is (-3/2,7/4)
II parabola
[tex]y=(x+1)^2[/tex]
Vertex =(-1,0)
Only I graph solves the system of equations
has vertex as
