Can someone tell me what the blue lines mean?
2 answers:
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In calculus, we use derivatives to find the instantaneous rate of change at any point on a graph. To find the average rate of change, we just find the slope of the secant line that intercepts two points on the graph.
We find slope with the following equation:
In this case, we are looking for the slope from x = -1 to x = 1. We have both x values, so next we need the y values.
F(-1) = (-1)^2 - (-1) - 1 = 1
F(1) = (1)^2 - (1) - 1 = -1
Now plug in the x and y values to find the slope:
The answer is -1.
We find slope with the following equation:
In this case, we are looking for the slope from x = -1 to x = 1. We have both x values, so next we need the y values.
F(-1) = (-1)^2 - (-1) - 1 = 1
F(1) = (1)^2 - (1) - 1 = -1
Now plug in the x and y values to find the slope:
The answer is -1.
Answer:
Step-by-step explanation:
focus at (-5, -3), and directrix y = -6
Directrix y=-6 so its a vertical parabola
so equation is
(x-h)^2 = 4p(y-k)
(h,k) is the center
P is the distance between focus and vertex
distance between focus and directrix = 2p
distance between -3 and y=-6 is 3
2p = 3
p = 3/2 or p = 1.5
Focus is (h, k+p)
given focus is (-5, -3) so h= -5 and k+p = -3
k+p=-3, plug in 1.5 for p
k + 1.5 = -3
subtract 1.5 on both sides
k = -4.5
(x-h)^2 = 4p(y-k)
divide by 6 on both sides
then subtract 4.5 on both sides
focus at (10, -4), and directrix y = 6.
Directrix y=6 so its a vertical parabola
so equation is
(x-h)^2 = 4p(y-k)
distance between focus and directrix = 2p
distance between -4 and y=6 is -4-6=-10
2p = -10
p = -5
Focus is (h, k+p)
given focus is (10, -4) so h= 10 and k+p = -4
k+p=-4, plug in 5 for p
k - 5 = -4
add 5 on both sides
k = 1
(x-h)^2 = 4p(y-k)
divide by -20 on both sides and add 1 on both sides