y=4√(x)
y=(1/16)x2
y=-∛(x)
y=-x3
Graph y=∛(x). Find f(-8)(-1)(1)(8)
(-8,-2),(-1,-1),(1,1),(8,2)
Factor 3st-6s+8t-16
(3s+8)(t-2)
Solve for x: y=ex
ln(y)=x
y=x2 + 4
y=-√(x-4)
y=2∛(x)
y=(1/8)x3
Graph y=2√-(x-1). Find f(0),(-3),(-8)
(0,2),(-3,4),(-8,6)
Find the roots of y=-x2+8x-12
x=2, x=6
Solve for x: y=8x
log8(y)=x
y=-√(-2(x))
y=-(1/2)x2
y=(1/125)x3 + 3
y=5∛(x-3)
Graph y=-√((1/2)x). Find f(2)(8)(18)
(2,-1),(8,-2),(18,-3)
Find the roots of y=15x3+45x2+30x
x=0, x=-2, x=-1
When is 2x ≥ 4000
When x ≈ 12.
y=√(-(1/2)(x)) + 2
y=-2x2+8x-8
y=-(x3+1)/2
y=∛(-2x-1)
Graph y=-(1/2)∛(x-1) + 2. Find f(-7)(1)(2)
(-7,3),(1,2),(2,3/2)
64x3 - 125
(4x-5)(16x2+20x+25)
A loan of $20,000 at 7% is due in 10 years. How much is due?
f(10)=$20,000(1+.07)x = $39,343.03.
y=5√(x+5) - 7
y=(x2+14x+49)/25 - 5
y=∛(x) + 1
y=x3-3x2+3x-1
Graph y=4√((1/2)(x-2)) + 2. Find f(0)(6)(16)
(0,2),(6,6),(16,10)
What is the end behavior of y=-x3-13x2-52x-60?
Left: y↦ +∞ ,x↦ -∞
Right: y↦ -∞ ,x↦ +∞
A new car ($66,000) loses 40% value in 5 years. How much is it worth after 2 years? Put in form y=a·ekx
y=66000eln(.903)(2) = $53,816.99