Domain of y = x2
(-∞, ∞)
Axis of Symmetry of y=−2(x+1)2+8
x=-1
Inverse of: f(x)=3x+5
f-1(x)= x - 5 / 3
Write the equation of a parabola with:
vertex (2,−4) , a=3
y=3(x−2)2−4
y=x2+6x+5
y=(x+3)2−4
Domain / Range of: x2+y2=16
Range[-4, 4]
Max / Min of: y=3(x−3)2−6
Inverse of f(x) = x2 + 8
f-1(x) = +- √x-8
A parabola has vertex (1,2) and passes through point (3,10). Write an equation.
y=2(x−1)2+2
y=2x2+8x+3
y=2(x+2)2−5
Domain / Range of: y=−√(x+2)+4
Vertex of y=-2(x+4)2+5
(-4, 5)
What is the 3-step process for finding the inverse of a function?
Write the inverse as y =
Swap x and y
Solve for y
A parabola has vertex (−2,3) and passes through (0,11). Write the equation.
y=2(x+2)^2+3
y=−3x2−12x−7
y=−3(x+2)2+5
Range of y=−2(x−1)2+8
Range: y≤8
End Behaviours of -3√(2x+4) +2.
x -> ∞ y -> -∞
X -> -2+ y -> 2
Will f-1(x) = +- √x-8 have EB's? Why / Why not?
No. Fails VLT. Not a function
Write an equation: A reflection in y axis, vertical stretch by 2, horizontal shrink by 1/3, horizontal shift left 6, vertical shift up 12
y=2(-3(x+6))2+12
y=−2x2+8x+5
y=−2(x−2)2+13
Domain / Range of: y=−3x2+12
DOP, Vertex, End Behaviours, AOS, Max / Min of: y=−2x2+12x−5
y=−2(x−3)2+13
Vertex: (3,13)
Down
AOS: x=3
So it has a maximum value of 13 when x = 3.
EB: As x→∞, y→−∞ , As x→−∞, y→−∞x
Find the inverse of: f(x)=(x−1)2
f-1(x)= √(x) +1
Write equation; Reflection in x , y axis, VC by 1/2, HC by 1/2, H. shift right 10 units, vertical shift down 7 units
y = -1/2 ( -2 (x-10))2-7
y = -2x2 - 3x + 7
y = -2 (x+3/4)2+65/8