70. pts2x + 6 ≥ –4

A.
x ≤ –5

B.
x ≥ –8

C.
x ≥ –5

D.
x ≤ –8

Answer:

Step-by-step 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

Answer:

f(2) = 3

Step-by-step explanation:

Substitute 2 for x, and evaluate the expression.

f(x) = 2x - 1

f(2) = 2(2) - 1

f(2) = 4 - 1

f(2) = 3