Rational
f(x) = 2x^2/x^2-4
Find the Vertical Asymptotes
x^2-4 = 0
x^2=4
x = 2 and -2
f(x) = x^2-6x+9
Find the x and y intercepts
x int: 3
y int: 9
x int: set f(x) = 0, solve for x
y int: set x = 0, solve for f(x)
f(x) = √x-4
Find the x intercept.
x int: 4
The -4 means the graph moves 4 spts to the right
f(x) = 3|x-5|+7
Find the vertex
(5, 7)
-5 inside the absolute value -> right 5
+7 outside the absolute value -> up 7
f(x) = 2x^2/x^2-4
Find the Domain and Range
D: (-infinity, -2)(-2, 2)(2, infinity)
R: (-infinity, 0)(2, infinity)
Explanation: For domain, find the vertical asymptotes. For range, find the horizontal asymptotes.
f(x) = x^2-8x+20
find the vertex
(4, 4)
To find the vertex, use -b/2a. This gives the x coordinate 4. Plug 4 in for x and solve for f(x).
f(x) = √(x+2)-6
Find the Domain and Range
D: [-2, infinity)
R: [-6, infinity)
Find the vertex and graph the function. -6 means the graph shift down 6 spots, and the +2 means the graph shifts left 2 spots.
f(x) = 3|x-5|+7
Find the x and y intercepts
x int: none
y int: 22
3|x-5|+7 = 0 -> |x-5| = -7/3
can't be negative absolute value
3|0-5|+7 = 22
f(x) = 4x+10/2x^2-8
Find when f(x)>0 and f(x)<0
f(x)>0: (-2.5, -2)(2, infinity)
f(x)>0: (-infinity, -2.5)(-2, 2)
Explanation: Find the x intercept (-2.5) and the HA and VA. The graph starts under the HA and goes above it at the x intercept.
f(x) = x^2-3x-10
When is f(x)>0? When is f(x)<0?
f(x)>0: (-infinity, -2)(5, infinity)
f(x)<0: (-2, 5)
Find the x intercepts. The graph is above 0 on both sides of the x intercepts, and below 0 between the x intercepts.
f(x) = √(x-7)-3
When is f(x)>0? When is f(x)<0?
f(x)>0: (16, infinity)
f(x)<0: [7, 16)
Find the vertex and the x intercept. From the vertex until the x intercept, the graph is under the x axis. After the x intercept, the graph is over the x axis.
f(x) = 3|x-5|+7
Find the Domain and Range
D: (-infinity, infinity)
R: (7, infinity)
Graph the equation. Domain is every real number. For range, there is nothing below the vertex(5, 7), and then every real number above it.
f(x) = x^2-4/x+3
find the slant asymptote
SA = x-3
Explanation: divide x^2-4 by -3
f(x) = x^2+8x+15
When is the graph increasing? When is it decreasing?
increasing: (-4, infinity)
decreasing: (-infinity, -4)
Find the vertex. The graph decreasing until the vertex, then increases after the vertex.
f(x) = -√(x-6)+5
When is f(x)>0? When is f(x)<0?
f(x)>0: [6, 31)
f(x)<0: (31, infinity)
Because of the negative, the graph is flipped. It starts over the x axis at the vertex, then is below the x axis after the vertex.
f(x) = 3|x-5|+7
When is the graph increasing? When is it decreasing?
increasing: (5, infinity)
decreasing: (-infinity, 5)
Graph the equation. it is decreasing until the vertex, then it starts increasing.
f(x) = x^2-11x+28
Find the Domain and Range
D: (-infinity, infinity)
R: (-2.25, infinity)
In a quadratic equation, the domain is all real numbers. Since the x^2 is positive, the range is from the vertex until all real positive numbers. The vertex is (5.5, -2.25)
https://www.desmos.com/calculator/tfrr9p4kte
Find the function for this graph
f(x) = √(x-6)+7
Look at where the vertex is. It is at the point (6, 7) which means it shifted 6 spots right and 7 spots up. A shift 6 spots right indicates -6 inside the parenthesis, and a shift 7 spots up indicates a +7 outside the parenthesis.
https://www.desmos.com/calculator/7ub32dsrza
Find the function
f(x) = 2|x-9|+4
The slope is 2. The -9 means that the graph shifts 9 spots to the right. The +4 means the graph shifts 4 spots up.