Given the function
q(x)=2(x+1)2-2
Will the graph of the parabola open up or down?
Up
Solve using the square root method.
x2-121=0
x=±11
A stone is thrown from the top of a cliff. The height od the stone can be modeled by h(t)=-4.9t2+10t+100 where t is time in seconds since the stone is released and h(t) is the height of the stone in meters.
What is the initial height of the stone?
100 meters
List the possible rational zeros of
z(x)=x4+7x2-3x-15
±1,±3,±5,±15
Use synthetic division to divide
(x3-7x2+13x-3)÷(x-3)
= x2-4x+1
Remainder 0
f(x)=-x4-2x3-5x2-29
Down, Down
Determine the vertex of the given function
g(x)=3(x+2)2-3
(-2,-3)
A stone is thrown from the top of a cliff. The height od the stone can be modeled by h(t)=-4.9t2+10t+100 where t is time in seconds since the stone is released and h(t) is the height of the stone in meters.
What is the height of the stone after 5 seconds?
27.5 meters
Given f(x)=3x3+7x2-28x-32
Factor f(x), given that -1 is a zero.
f(x)=(x+1)(x+4)(3x-8)
DAILY DOUBLE! Choose any number up to 500!
Divide using synthetic division
(7x2+6x+10)÷(x+7)
=7x-43+311/x+7
Remainder 311
DAILY DOUBLE! Choose any number up to 500!
If the leading coefficient of a function is negative and the degree is odd, what will the end behavior of the graph of the function be?
Up, Down
Determine the axis of symmetry of the function
k(x)=3(x+3)2-3
x=-3
A long jumper leaves the ground at an angle of 21°above the horizontal, at a speed of 9/ms. The height of the jumper can be modeled by h(x)=−0.049x2+0.365x,where h is the jumper's height in meters and x is the horizontal distance from the point of launch.
At what horizontal distance from the point of launch does the maximum height occur? Round to 2 decimal places.
approximately 3.72 meters
Given f(x)=3x3+7x2-28x-32
Solve f(x)=0. Express your answers in simplest form.
{-1, -4, 8/3}
Use long division to divide
(3x3-11x2-32)÷(x-2)
=3x2-5x-10-52/x-2
Remainder 52
Determine the end behavior of the graph of the function:
k(x)=11x7+9x2+9x+6
Down, Up
Find the vertex of the parabola using the vertex formula
g(x)=3x2-6x+112
(1,109)
A long jumper leaves the ground at an angle of 21°above the horizontal, at a speed of 9/ms. The height of the jumper can be modeled by h(x)=−0.049x2+0.365x,where h is the jumper's height in meters and x is the horizontal distance from the point of launch.
What is the maximum height of the long jumper? Round to 2 decimal places.
approx. 0.68 meters
DAILY DOUBLE! Choose any number up to 500!
Find all the zeros. Write your answer in exact form.
f(x)=x3-7x2+13x-3
The zeros of f(x) are
3, 2+√3, 2-√3
Use synthetic division to divide the polynomials
(2x2+26x+16)÷(x+7)
=2x+12 - 68/x+7
Remainder 68
State the end behavior and degree of the polynomial function.
f(x)=-(x+1)3(x-2)(x+4)2
Degree 6
End behavior: Down, Down
Write the domain and range of the function in interval notation
h(x)=3(x+2)2-3
domain (-inf, inf)
range [-3, inf)
A long jumper leaves the ground at an angle of 21°above the horizontal, at a speed of 9/ms. The height of the jumper can be modeled by h(x)=−0.049x2+0.365x,where h is the jumper's height in meters and x is the horizontal distance from the point of launch.
What is the length of the jump? Round to 1 decimal place.
approx 7.4 meters
Find the zeros of the function and state the multiplicities.
z(x)=x3+3x2-4x-12
(Hint:Use factor by grouping)
zeros : x=-3,x= 2, x= -2
each has a multiplicity of 1
Find all the zeros
f(x)=x4-12x3+82x2+36x-255
given that 6+7i is a zero
Write your answer as a solution set
{6±7i,±√3}