Question

1) Find the equation of the following lines:

A)Through the point (-4,2) and perpendicular to the line y=-1/2x +1

y-y1=m(x-x1)
The reciprical of -1/2 =2
So my answer is y-2=2(x+4) or simplified y=2x+10

B)Through the point (1,1) and parallel to the line 5y=15x+25

so y-1=15(x-1) or y+15x-14

Was I suppose to divide the 5x on both sides first?

2) Determine if the limit exists and compute if it does:

lim x->3 of x^2-x-6 / x-3

(x+2)(x-3) / x-3
The (x-3) on top and bottom cancel

(3+2) = 5

So the limit is 5

3) Find the first and second derivative of the following:

a) y=pi^2x

First= 2pi
Second=2

b) y=3x^6+2x^2

First=18x^5+4x
Second=90x^4+4

Are my answers to these questions correct? Thank you for you help!

Answer 1

all look ok, except 3 a) and 1 B)

in 1 B) divide your original by 5 to get
y = 3x + 5, so the slope is 3
new equation: y = 3x + b
sub in point (1,1,)
1 = 3+b
b = -2
new equation: y = 3x - 2

3 a) is your equation y = (π^2)x or y = π^(2x) ?
I will assume y = (π^2)x
your first derivative is π^2
but π^2 is a constant, so the 2nd derivative would be 0.

Answer 2

yes, 3b is (pi^2)x. I do not understand why the first derivative is pi^2.

Answer 3

π^2 is a constant
just like the derivative of 5x would be 5
so the derivative of (π^2)x is simply π^2

(think of (π^2) as 3.14^2 or appr. 9.86
so what would be the derivative of 9.86x ? )

Answer 4

ok thank you