Answer:
Explanation:
The given y=x²+4x+3 should be re-written in "vertex form," which is
y = a(x - h) + k, where (h, k) is the vertex.
Here y=x²+4x+3 = y=x²+4x + 4 - 4 +3 (completing the square), or
y = x^2 + 4x + 4 - 1. which condenses to y = (x + 2)^2 - 1
Thus, h = -2 and k = 1, and so the vertex is at (h, k), or (-2, 1). This parabola opens up. The axis of symmetry is the vertical line x = -2.
An alternative equation for a vertical parabola that opens up is:
y - k = 4p(x - h)^2. Rewriting y = (x + 2)^2 - 1 to fit this form leads to:
y + 1 = 4p(x + 2)^2. We must find the value of p that makes this equation true at any (x, y) on the graph. Suppose we arbitrarily choose x = 1. Then, according to y=x²+4x+3, y = 1^2 + 4(1) + 3, or y = 8 when x = 1.
Then y - k = 4p(x - h)^2 becomes 8 - 1 = 4p(1 + 2)^2, or
7 = 4p(9), or p = 7/36
The focus is thus 7/36 units above the vertex: (-2 + 7/36), and the directrix is the horizontal line which is 7/36 units below the vertex: y = -2 - 7/36.
Vertex: (-2, 1)
Focus: (-2, 2 7/36)
Directrix: y = -2 7/36