Determine if f(x)=x²+3x+2 is a function.
Yes, it passes the vertical line test.
State the domain of f(x)=x²+3x+2.
Domain: all real numbers
State the range of f(x)=x²+3x+2.
Range: y ≥ -0.25
(vertex y-value)
Complete the square:
f(x)=x²+3x+2
Think divide by 2, then square
f(x)=(x²+3x + 1.5² - 1.5²) +2
f(x)=(x²+3x + 2.25) + 2 - 2.25
f(x)=(x²+3x + 2.25) - 0.25
f(x)=(x+1.5)² - 0.25
Vertex = (-1.5, -0.25)
Parabola end behavior is quadrant 2 to quadrant 1 (smile), so we have a minimum.
Range
Interval notation: [-0.25, infinity)
Set notation: {y|yER, y >= -0.25}
Graphing calculator:
Enter the function in Y1: .
Press GRAPH to see the parabola.
Press 2nd → TRACE → 3: minimum.
Move the cursor near the bottom of the parabola. The calculator displays the vertex coordinates automatically.
Tip: You can also check the value by using the TABLE feature near X = -1.5 to see that .
Find f(4) for f(x)=x²-5x+6.
f(x)=x² - 5x + 6
f(4)= 4² - 5(4) + 6
f(4)= 16 - 20 + 6
f(4) = 2
Apply transformation: y=(x-3)²+5 — what is the vertex?
Vertex: (3, 5)
Apply transformation: y=-2|x+4| — state reflection.
Reflection over x-axis
Horizontal shift: → shifts the graph 4 units left.
Vertical stretch: in front → stretch vertically by a factor of 2.
Reflection: The negative sign reflects the graph over the x-axis, so the “V” opens downward.
Enter in Y1.
Press GRAPH to visualize.
Use 2nd → TRACE → maximum to find the vertex if needed (since it’s reflected, the vertex is now the maximum).
Summary: Reflection is over the x-axis.
Determine if relation {(1,3),(2,5),(3,5)} is a function.
Yes, each x has exactly one y.
Determine if relation {(1,3),(1,4)} is a function.
No, x=1 maps to two different y-values.
Find (f∘g)(x) for f(x)=x+1, g(x)=2x.
(f∘g)(x)=2x+1
Find inverse of f(x)=2x-6.
f⁻¹(x)=(x+6)/2
Solve for x if f(x)=x²+2x, f(x)=0.
x=0 or x=-2
Identify vertical stretch factor in y=3(x-2)².
Vertical stretch factor = 3