Download Review: Polynomial Functions

Review: Polynomial Functions
(1st topic covered; Chapter 2 in text)
1. For each of the following functions, state the degree, and all roots in
simplest radical form if possible. Then make a rough sketch. Location of
local max/min. points are not necessary.
a) f ( x)  ( x  4)2 ( x 2  6 x  9)
b) f ( x)  ( x 2  9)( x 2  4 x  3)
c) f ( x)  x3  4 x 2  x  4
a) quartic function (4th degree).
f(x) = - (x+4)(x+4)(x+3)(x+3)
So the roots are x = -4, -4, -3, -3
Sketch:
b) quartic function
f(x) = (x+3)(x-3)(x+3)(x+1)
So the roots are x = -3, -3, -3, -1
Sketch:
c) cubic function
f(x) = (x-1)(x-4)(x+1) (Use synthetic division to get these factors)
So the roots are x = 1, 4, -1
Sketch:
2. Solve using a sign chart. Use interval notation.
x 3  7 x 2  4 x  28  0
What does answer represent, graphically?
Find the roots and do a sign chart. Using synthetic division, the factors are:
f(x) = (x-2)(x2 – 5x – 14) = (x-2)(x-7)(x+2)
So the roots are x = 2, 7, -2
The sign chart is
- -2
+
2
7
+
The final answer is x ε (-2, 2] U [7, ∞)
It represents where the graph of y = x3 – 7x2 – 4x + 28 is above the x –axis.
3. List all possible rational roots using the rational root theorem. Then factor
to find all roots.
a) g ( x)  x 4  7 x 2  12
b) h( x)  x3  x 2  16 x  20
a) Possible rational roots are ±1, 2, 3, 4, 6, 12
Roots are x = 2, 2, -2, -2 (Use synthetic division or “quadratic substitution”…
ie. let x2 = a and solve that equation).
b) Possible rational roots are ±1,2,4,5,10,20
Roots are x = -5, 2, 2 (Use synthetic division)
4. A cubic function goes through the following points:
Find the equation of the cubic function that fits the data. You may
x
y
use substitution, elimination, or inverse matrices.
0
0
1
-12
Since y = ax3 + bx2 + cx + d, we have:
3
6
6
48
(0, 0) gives 0 = (0)a + (0)b + (0)c + d or in other words, d = 0
(1, -12) gives –12 = (1)a + (1)b + (1)c + d or –12 = a + b + c
(3, 6) gives 6 = (3)3a + (3)2b + 3c + d or 6 = 27a + 9b + 3c
(6, 48) gives 48 = (6)3a + (6)2b + (6)c + d or 48 = 216a + 36b + 6c
Using one of the three methods listed above, we find that a = -1, b = 11,
c = -22 and we know already that d = 0. So the final equation is:
y = -x3 + 11x2 – 22x
5. If f(x) = 3x + 5, g(x) = x2 and h(x) = 2x, find
a) f(g(2))
b) h(g(x))
c) f(g(h(1)))
a) f ( g (2))  3((2) 2 )  5  17
b) h( g ( x))  2 x 2


c) f ( g (h(1)))  3 2(1)   5  17
2
6. Find the equation of the tangent to y  x3  4 x  10 at x = -1. Remember to
express the answer in the form y = mx+b.
The slope of the tangent is y’ = 3x2 + 4 using the power rule.
At x = -1, this equals 7. That represents a positive and pretty steep slope!
When x = -1, y = (-1)3 + 4(-1) + 10 = 5
I used the ORIGINAL function to find this y value, because I am finding a
point on the curve. Its ordered pair is (-1, 5)
So in y = mx + b, we know that m = 7, x = -1 and y = 5. Find b!
5 = 7(-1) + b
b = 12
So the answer is y = 7x + 12
Graphically, here’s what the equation of the tangent line looks like against the
original function:
7. Given f ( x)  x3  12 x , sketch an accurate graph showing the roots (exact),
location of local max./min. points (you will have to find f’(x) and set it equal
to 0 to do this) Then state the domain, range, intervals of increase and
decrease and intervals where the function is above and below the x-axis.
To get the roots, factor.
f(x) = x(x2 – 12)
0 = x(x2 – 12)
x = 0, x = ±√12 (this is x = ±2√3 in simplest radical form, approx. ±3.46)
To get the local max./min. points, find f’(x). The slope will be 0 at the max/min,
so set f’(x) = 0 and solve for x.
f’(x) = 3x2 – 12
0 = 3x2 – 12
0 = 3(x2 – 4)
x = ±2
When x = 2, y = -16 in the original function.
When x = -2, y = 16 in the original function.
So the local max. is at (-2, 16) and the local min. is at (2, -16)
Sketch:
Analysis:
D: x ε R
R: y ε R
Inc: x ε (-∞, -2) U (2, ∞)
Dec: x ε (-2, 2)
Above x-axis (f(x) > 0): x ε (-2√3, 0) U (2√3, ∞)
Below x-axis (f(x) < 0): x ε (-∞, -2√3) U (0, 2√3)
8. Given that y  x 2  3x +5,
a) Find y’ using the power rule.
b) Find y’ using the definition of the derivative, ie. limits:
f ( x  h)  f ( x )
y '  lim h  0
h
c) State the equation of the tangent line to this function at (3, 20)
a) y’ = 2x + 3
b)
( x  h) 2  3( x  h)  5  ( x 2  3 x  5)
h
2
2
x  2hx  h  3 x  3h  5  x 2  3x  5
 lim h0
h
2
2hx  h  3h
 lim h0
h
h(2 x  h  3)
 lim h0
h
 lim h0 2 x  h  3
y '  lim h0
 2x  3
c) y = mx + b where x = 3, y = 20 and m = 9 (m = 2x + 3 where x = 3)
So b must equal –7.
The final answer is y = 9x – 7
1
x  1,  ,  5 and the function
3
has a y-intercept of . State the equation of this function in general form.
9. Suppose the roots of a quartic function are
In factored form, the equation of this function would be
f ( x)  k ( x  1)(3x  1)( x 2  5)
Notice how I changed all the given roots into their corresponding factors such
that f(x) = 0 to find the roots. Note the “k” out front… constants factored out don’t
affect the roots. But they can affect the y-intercept.
Our y-intercept is 25. Substituting x = 0 to obtain the y-intercept, we have


k  0  1 3(0)  1 0 2  5  25
k (1)(1)(5)  25
5k  25
k 5
Now expand the factored form to get it in general form.