Asked by idkhowtodomath
Can anyone use these two questions as an example,I will do the rest,thank you!
1) y=x^2-3x, for -2≤x≤5.
(Scales 2cm for x, 1cm for y)
Find:
a) gradient of the tangent to the curve at x=3.
b) gradient of the tangent to the curve at x= -1.
c) value of x where the gradient of the curve is zero.
2) The number of an atoms of a particular chemical element is modeled by the following function: y=600x3^-x where x is the number of hours.
a) How many atoms of the elements were there at the start?
b) Draw a graph for 0≤x≤4.
c) At what time were there 150 atoms?
1) y=x^2-3x, for -2≤x≤5.
(Scales 2cm for x, 1cm for y)
Find:
a) gradient of the tangent to the curve at x=3.
b) gradient of the tangent to the curve at x= -1.
c) value of x where the gradient of the curve is zero.
2) The number of an atoms of a particular chemical element is modeled by the following function: y=600x3^-x where x is the number of hours.
a) How many atoms of the elements were there at the start?
b) Draw a graph for 0≤x≤4.
c) At what time were there 150 atoms?
Answers
Answered by
oobleck
#1. y = x^2-3x
y' = 2x-3
(a) plug in x=3
(b) plug in x = -1
(c) 2x-3 = 0
#2. You ought to use * for multiplication, so it does not get confused with x the variable
y = 600 * 3^-x
(a) plug in x=0
(b) surely by now you know the shape of exponential curves. google can help if not.
(c) solve 600 * 3^-x = 150
3^-x = 1/4
3^x = 4
x log3 = log4
x = log4/log3
y' = 2x-3
(a) plug in x=3
(b) plug in x = -1
(c) 2x-3 = 0
#2. You ought to use * for multiplication, so it does not get confused with x the variable
y = 600 * 3^-x
(a) plug in x=0
(b) surely by now you know the shape of exponential curves. google can help if not.
(c) solve 600 * 3^-x = 150
3^-x = 1/4
3^x = 4
x log3 = log4
x = log4/log3
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