Inverses:
To take the inverse of any function f(x), you:
(1) write down the function,
(2) switch the x and the y, and
(3) solve for y.
The graph of the inverse is the reflection of the original function's graph over the line y=x.
example: y = 3x - 5. inverse: x = 3y - 5; x + 5 = 3y; y = (x+5)/3.
Compositions:
To take the composition f(g(x)), you:
(1) write down the function f(x), but everywhere there's an x, put an empty parentheses: ( ), leaving some space, and then
(2) write in g(x) in all of the empty parentheses.
example: f(x) = 2x + x
2 – 1, g(x) = 7x + 1/x.
f(
) = 2(
) + (
)
2 – 1;
f(g(x)) = 2(7x + 1/x) + (7x + 1/x)
2 – 1
g(
) = 7(
) + 1/(
);
g(f(x)) = 7(2x + x
2 – 1) + 1/(2x + x
2 – 1)
Quadratics, completing the square:
To complete the square in y = ax
2+bx+c, you:
(1) group all the x terms together in a parentheses and leave a space,
(2) if there is a number or a negative sign in front of the x
2 term (let's call that number "A"), take it out of the parentheses by dividing all x terms by it,
(3) write down, off to the side, B = half of the number in front of x, and C = B
2,
(4) add C inside the parentheses, and subtract C times the number that was originally in front of x
2 at the end of the function (C*A), to keep it balanced,
(5) rewrite the function as y=A(x+B)
2 + some constant. Now you know the transformations that have been done on y=x
2: stretching vertically by A, moved left by B, and moved up by the constant value.
example: y = 3x
2 + 30x + 71
y = (3x
2 + 30x
) + 71
[A=3]
y = 3(x
2 + 10x
) + 71
[B=½*10=5; C=5
2=25]
y = 3(x
2 + 10x + 25) + 71 – 3*25
y = 3(x + 5)
2 – 4
(stretched vertically by 3, moved 5 left, moved 4 down, so vertex is at (-5,-4))
Quadratics, finding the roots:
Once you’ve completed the square, you find the roots by setting y=0 and solving for x.
example: y = 3(x + 5)
2 – 4
3(x + 5)
2 – 4 = 0
3(x + 5)
2 = 4
(x + 5)
2 = 4/3
(x + 5) = ±sqrt(4/3)
x =
- 5 ±sqrt(4/3)
(sqrt = square root)
Quadratics, quadratic formula:
Let’s try completing the square on ax
2+bx+c = 0:
(ax
2+bx
) + c = 0
a(x
2 + (b/a)x
) + c = 0
[B = ½ (b/a) = b/(2a); C = B
2 = b
2/(4a
2)]
a(x
2 + (b/a)x + b
2/(4a
2)) + c - b
2/(4a) = 0
a(x + b/(2a))
2 + (4ac – b
2)/(4a) = 0
Now solve for x:
a(x + b/(2a))
2 = (b
2 – 4ac)/(4a)
(x + b/(2a))
2 = (b
2 – 4ac)/(4a
2)
(x + b/(2a)) = ±(sqrt(b
2 – 4ac))/(2a)
x = [-b ± sqrt(b
2 – 4ac)]/(2a)
This is known as the “quadratic formula.” If you can remember this formula, then you can always just plug in your numbers for a, b, and c to find the roots of ANY quadratic function.
Cubics: factoring by grouping
For SOME cubic functions, you can use a method called factoring by grouping to find the roots. This does not work for all cubics. To do this,
(1) write the function in order with the highest order terms first, decreasing down to the constant term,
(2) factor just the first two terms by themselves and the last two terms by themselves, and
(3) see if there is a common factor between the first two terms and the last two terms – if there is, that can be factored out of the entire function.
example: y = 12x
3 – 20x
2 – 27x + 45
y = [12x
3 – 20x
2] + [-27x + 45]
y = [4x
2(3x – 5)] + [-9(3x – 5)]
y = (3x – 5)(4x
2 – 9)
If the constant term in the quadratic part is NEGATIVE, then you can factor it, too:
example: y = (3x – 5)(4x
2 – 9)
y = (3x – 5)(2x + 3)(2x – 3),
which has roots at x = 5/3, -3/2, and 3/2.
