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Find the distance and slope
(5, -4) and (-1, 1)

Find maximum or minimum of the function
y= 3-5x +x^2

2007-12-15 11:44:07 · 5 answers · asked by jamie d 2 in Science & Mathematics Mathematics

5 answers

slope

(y2-y1)/(x2 - x1)
(-4-1) /(5-(-1))= -5/6

distance
sqrt(61);

take the derivative to find the velocity
when velocity is 0 that is you min or max.
take it again that would tell you your acceleration if it is concave up or down

p(x) = x^2-5x +3
v(x) = 2x -5 solve for x = 2.5
a(x) = 2

so it's concave up and it is your minimum at (2.5,-3.25).

2007-12-15 12:06:02 · answer #1 · answered by iknowu2jan 3 · 0 0

idk about the distance but the slope is y2-y1 divided by x2-x1
which is 1+4 divided by -1-5...which equals 5/-6 which is the slope.

this would be a minimum since the function is going up.
x^2-5x+3

-b/2a

5/2= 2.5 which is the x axis

then plug in 2.5 for x

you should get -3.25 and thats the y


so minimum is (2.5, -3.25)

you welcome

2007-12-15 11:58:02 · answer #2 · answered by Anonymous · 0 0

distance = sqrt((5+1)^2 + (-4-1)^2) = sqrt(61)

slope = (1--4)/(-1-5)=-5/6

vertex = -b/2a = 5/2
y(5/2) = 3 - 5*5/2 + (5/2)^2
= -3.25

2007-12-15 12:02:03 · answer #3 · answered by norman 7 · 0 0

slope = 1 - -4 over -1 - 5
you will get -5/6

the distance = the squer rote of (-1-5)^2 + ( 1--4)^2
you will get -6^2 + 5^2 = -36 + 25
the answer you will get is the squer rote of -11

2007-12-15 14:38:21 · answer #4 · answered by $$$$$$$$$$$$$ 2 · 0 0

Distance:
the square root of 61. (you might want to see if you can simplify that.)

Slope:
-4/3

Min:
(2.5,-3)

2007-12-15 11:51:30 · answer #5 · answered by Anonymous · 0 0

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