Let's examine the given function first:
f(x) = x^2 + 1 is the same as f(x) = 1(x-0)^2 + 1.
The vertex of the graph of this function is at (0, 1).
Let x=0 to find the y-intercept: f(0)=0^2+1 = 1; y-int. is at (0,1) (which happens to be the vertex also)
Comparing f(x) = x^2 + 1 to y = x^2, we see that the only difference is that f(x) has a vertical offset of 1. So: Graph y=x^2. Then translate the whole graph UP by 1 unit. That's it. Note (again) that the vertex will be at (0,1), and (0,1) is also the y-intercept.
1/32 maybe................
Answer:
A
Step-by-step explanation:
slope is (y2-y1)/(x2-x1)
anyhow that comes out to -3/2
the only place that actually applies is in A
First, lets transform the given vector into an unit vector (dividing by its module)
UnitVec = 4/5 i + 3/5 j
Then lets change this vector into a polar form
UnitVec = 1. with angle of 36.869 degrees taking as a reference the i vector
Then, the probem tells us that the vectors u and v make an angle of 45 degrees with UnitVec, so lets add+-45 to the vector in polar form
U = 1*[cos(36.869 +45)i + sin(36.869 +45)j] = 0.1414 i + 0.9899 j
V = 1*[cos(36.869 -45)i + sin(36.869 -45)j] = 0.9899 i - 0.1414 j
