G
Guest
Guest
So i have a couple questions..well three.
1. Find the coordinates of the points on the graph of y=x³+2 at which
the slope of this tangent is 12.
y=x³+2
= 3x^2, slope is 12
Therefore, x much equal -2 and 2 in order for the slope to be 12. But im
unsure on how to find the y's.
2. Find the values of x so that the tangent to the function
y=3
----
³√x is parallel to the line x+16y+3=0.
y=3
---
³√x
= 3x^1/3
=-x^4/3
Im not sure how to find the slope of the line and i know if i had that i
could sub that into -x^4/3.
3. Find the slope of the tangent. y=2/x (-2, -1)
y=2/x (-2, -1)
=2x^-1
=2(-1/2)x^-2
=-x^-2
=-1
----
x^2
=-1
----
4
Im not sure if that is right or not, i doesnt look right to me.
Thanks for any help and i apologize if it is hard to understand, just tell
me and i will try my hardest to fix it.
1. Find the coordinates of the points on the graph of y=x³+2 at which
the slope of this tangent is 12.
y=x³+2
= 3x^2, slope is 12
Therefore, x much equal -2 and 2 in order for the slope to be 12. But im
unsure on how to find the y's.
2. Find the values of x so that the tangent to the function
y=3
----
³√x is parallel to the line x+16y+3=0.
y=3
---
³√x
= 3x^1/3
=-x^4/3
Im not sure how to find the slope of the line and i know if i had that i
could sub that into -x^4/3.
3. Find the slope of the tangent. y=2/x (-2, -1)
y=2/x (-2, -1)
=2x^-1
=2(-1/2)x^-2
=-x^-2
=-1
----
x^2
=-1
----
4
Im not sure if that is right or not, i doesnt look right to me.
Thanks for any help and i apologize if it is hard to understand, just tell
me and i will try my hardest to fix it.