Answer:
[tex]f(x)=x^2+3x+2[/tex]
Step-by-step explanation:
step 1
we have the points
(-1,0), (-2,0), and (0,2)
Plot the points
using a graphing tool
see the attached figure
The graph of a quadratic function must be a vertical parabola open upward
The vertex is a minimum
The quadratic function in general form is equal to
[tex]f(x)=ax^2+bx+c[/tex]
Substitute the value of x and the value of y of each given ordered pair in the general equation and solve for a,b and c
(0,2)
For x=0, y=2
substitute
[tex]2=a(0)^2+b(0)+c[/tex]
[tex]c=2[/tex]
(-1,0)
For x=-1, y=0
substitute
[tex]0=a(-1)^2+b(-1)+2[/tex]
[tex]0=a-b+2[/tex] ----> [tex]a=b-2[/tex] ----> equation A
(-2,0)
For x=-2, y=0
substitute
[tex]0=a(-2)^2+b(-2)+2[/tex]
[tex]0=4a-2b+2[/tex] ----> equation B
we have the system
[tex]a=b-2[/tex] ----> equation A
[tex]0=4a-2b+2[/tex] ----> equation B
substitute equation A in equation B
[tex]0=4(b-2)-2b+2[/tex]
solve for b
[tex]0=4b-8-2b+2[/tex]
[tex]2b=6[/tex]
[tex]b=3[/tex]
Find the value of a
[tex]a=3-2=1[/tex]
therefore
The quadratic function in general form is equal to
[tex]f(x)=x^2+3x+2[/tex]
see the attached figure N 2 to better understand the problem
