Let's take a look at D:
<span>D) y = (x-1)^2 - 16 Compare this to
y = (x-h)^2 + k This is the std. equation of a parabola in vertex form.
You can see, by comparison, that h=1 and k= -16; these are the coordinates of the vertex, clearly shown in the diagram.
Since the coefficient of (x-h)^2 is +1, the graph opens upward (which the given graph confirms), and is neither compressed nor stretched vertically.</span>
To solve for this, we need to find for the value of x
when the 1st derivative of the equation is equal to zero (or at the
extrema point).
So what we have to do first is to derive the given
equation:
f (x) = x^2 + 4 x – 31
Taking the first derivative f’ (x):
f’ (x) = 2 x + 4
Setting f’ (x) = 0 and find for x:
2 x + 4 = 0
x = - 2
Therefore the value of a is:
a = f (-2)
a = (-2)^2 + 4 (-2) – 31
a = 4 – 8 – 31
a = - 35
Do distributed property
7/3+3(2/3-1/3)^2
7/3+3×(2/3-1/3)^2
7/3+3×(1/3)^2
7/3×3×1/9
7/3×1/3
7+1/3
8/3 or 2 and 2/3 or 2.66667
It is most likely 14 units so it would be A. I think.
His elevation is higher than 1600, where there is less oxygen and, air which technically contains oxygen, and the fact that elevation means going “up”.
