Given quadratic function [tex]f(x)= x^2+4x+3.[/tex]
Let us find vertex of the graph first.
x-coordinate of the vertex = -b/2a = -4/2(1) = -4/2 = -2.
y-coordinate of the vertex = (-2)^2+4(-2)+3 = 4-8+3 = -1.
Therefore, vertex is at (-2,-1).
The leading coefficient of quadratic function f(x)= x^2+4x+3 is 1.
And it's a positive number.
So, the parabola open up and it would have a minimum at it's vertex point.
Let us find some more points on the graph :
x=-3
(-3)^2 +4(-3) +3 = 9 -12 +3 =0
x=-1
(-1)^2 +4(-1) +3 = 1 -4 +3 = 0.
From the graph, we can see vertex is at (-2,-1) and x-intercepts at (-1,0), (-3,0) and y-intercept at (0,3).
Therefore, following is the key feature of the graph is given incorrectly:
B) y-intercept (0, 2)
