Question

20 points!

f(x)=x^2+6x+8

I need to find two seperate zeros which are a smaller x and a larger x, then the vertex of the parabola which is demonstrated by the equation. Any help is greatly appreciated, but please give the answer with an explanation as well!

Answer 1

Basically, to solve this, we need to use the equation:
x^2 + 6x + 8 = 0
Now, to do this we need to simplify:
(x + 4)(x + 2) = 0
(When you multiply these two, you get x^2 + 6x + 8, you can check)
To make 0, one of these two have to make 0. In this case, one could be -4, and one could be -2
So, x = -4, and x = -2 (-2 is larger, -4 is smaller.)
Knowing this, we need to use -b/2a to find the vertex:
-6/2 = -3
Knowing that x is -3, we input it:
-9 + 18 - 8 = 1
Knowing this, we can say that the vertex is (-3, 1)

Answer 2

You can factor it by considering factors of 8 that add to 6.
8 = 1·8 = 2·4
The latter pair add to 6, so your factorization is
f(x) = (x + 2)(x + 4)
This product is zero when one of the factors is zero. There are 2 factors, hence 2 zeros of the function.
For x+2 = 0
x = -2
For x+4 = 0
x = -4

The graph of a parabola is symmetrical about its vertex, so the line of symmetry is halfway between the zeros, at x = (-2-4)/2 = -3. The y-coordinate of the vertex is the value of the function at this point,
f(-3) = (-3+2)(-3+4) = -1

a) Your two separate zeros are
x = -4 (smaller value)
x = -2 (larger value)

b) The vertex of the parabola is
vertex = (-3, -1)

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You can use a graphing calculator to answer these questions directly from the graph (or to check your work, as the case may be).