Which graph represents the function f(x) = x2 + 3x + 2?
graph 1
graph 2
graph 3
graph 4
![Which graph represents the function fx x2 3x 2 graph 1 graph 2 graph 3 graph 4 class=](https://us-static.z-dn.net/files/d45/27047a3356f81526d85357eeddfe5c83.png)
![Which graph represents the function fx x2 3x 2 graph 1 graph 2 graph 3 graph 4 class=](https://us-static.z-dn.net/files/d0c/b770a4e2da06acff36d09a6f12e47bd6.png)
![Which graph represents the function fx x2 3x 2 graph 1 graph 2 graph 3 graph 4 class=](https://us-static.z-dn.net/files/d98/115965d0f964886cdb9991c8c68c6585.png)
![Which graph represents the function fx x2 3x 2 graph 1 graph 2 graph 3 graph 4 class=](https://us-static.z-dn.net/files/d13/96612b6a596533fbc4bef0f0db87122a.png)
Answer:
Graph 1
Step-by-step explanation:
Here, the given equation is,
[tex]f(x)=x^2+3x+2-----(1)[/tex]
For x-intercept, f(x) = 0
[tex]x^2+3x+2=0[/tex]
[tex]x^2+2x+x+2=0[/tex]
[tex]x(x+2)+1(x+2)=0[/tex]
[tex](x+1)(x+2)=0[/tex]
[tex]\implies x=-1\text{ or } -2[/tex]
So, the x-intercept of the function are (-1,0) and (-2,0)
Since, the line must has at least one x-intercept.
⇒ Graph 2 and Graph 4 can not be the graph of the given function,
Also, for y-intercept,
Put x = 0 in equation (1),
We get, f(x) = 2,
Hence, the y-intercept of the given function is (0,2),
But in Graph 3 the y-intercept of the function = (0,1)
⇒ Graph 3 can not be the graph of the given function,
Therefore, Graph 1 is the correct graph of the given function.