Graph each points:
f(x) = -13x + 1
when x = 0, y = 1
f(0) = -13(0) + 1
f(0) = 0 + 1
f(0) = 1
when x = 1, y = -14
f(1) = -13(1) + 1
f(1) = -13 + 1
f(1) = -14
when x = 2, y = -25
f(2) = -13(2) + 1
f(2) = -26 + 1
f(2) = -25
etc.
Graph each point and connect them.
1)16^2 or 4^4
2)3^2
3)4^2
4)5^2
5)15^2
6)14^2
4-70=7-6x
-64-7=6x
-71=-6x
71/6=x
Answer:
Vertical A @ x=3 and x=1
Horizontal A nowhere since degree on top is higher than degree on bottom
Slant A @ y=x-1
Step-by-step explanation:
I'm going to look for vertical first:
I'm going to factor the bottom first: (x-3)(x-1)
So we have possible vertical asymptotes at x=3 and at x=1
To check I'm going to see if (x-3) is a factor of the top by plugging in 3 and seeing if I receive 0 (If I receive 0 then x=3 gives me a hole)
3^3-5(3)^2+4(3)-25=-31 so it isn't a factor of the top so you have a vertical asymptote at x=3
Let's check x=1
1^3-5(1)^2+4(1)-25=-25 so we have a vertical asymptote at x=1 also
There is no horizontal asymptote because degree of top is bigger than degree of bottom
There is a slant asympote because the degree of top is one more than degree of bottom (We can find this by doing long division)
x -1
--------------------------------------------------
x^2-4x+3 | x^3-5x^2+4x-25
- ( x^3-4x^2+3x)
--------------------------------
-x^2 +x -25
- (-x^2+4x-3)
---------------------
-3x-22
So the slant asymptote is to x-1
