When we try to solve for f(-2), we are trying to find the range of the point that has a domain of -2. On the graph, when x equals -2, f(x) equals 2. The blank is 2.
We know that
X²+y²=9 -------> X²+y²=3²
is the equation of a circle with center (0,0) and radius r=3 units
so
<span>the translation of four units to the right and three units down is equals to move the center (0,0)--------> (0+4,0-3)------> (4.-3)
the new center of the circle is (4,-3)
the new equation is
(x-4)</span>²+(y+3)²=3²
see the attached figure
60 = a * (-30)^2
a = 1/15
So y = (1/15)x^2
abc)
The derivative of this function is 2x/15. This is the slope of a tangent at that point.
If Linda lets go at some point along the parabola with coordinates (t, t^2 / 15), then she will travel along a line that was TANGENT to the parabola at that point.
Since that line has slope 2t/15, we can determine equation of line using point-slope formula:
y = m(x-x0) + y0
y = 2t/15 * (x - t) + (1/15)t^2
Plug in the x-coordinate "t" that was given for any point.
d)
We are looking for some x-coordinate "t" of a point on the parabola that holds the tangent line that passes through the dock at point (30, 30).
So, use our equation for a general tangent picked at point (t, t^2 / 15):
y = 2t/15 * (x - t) + (1/15)t^2
And plug in the condition that it must satisfy x=30, y=30.
30 = 2t/15 * (30 - t) + (1/15)t^2
t = 30 ± 2√15 = 8.79 or 51.21
The larger solution does in fact work for a tangent that passes through the dock, but it's not important for us because she would have to travel in reverse to get to the dock from that point.
So the only solution is she needs to let go x = 8.79 m east and y = 5.15 m north of the vertex.
Vectors and scalars are not the same quantity
