help with algebra 2 (factoring).?

Question

Suppose a quadratic function f(x) = x^2 + bx + c has zeros p and q.

a) Let k be an integer. Using only b, c, and k (and x), write a new qudratic function whose zeros are p+k and q+k.

b) Repeat part (a), but write a new function whose zeros are p-k and q-k.






Let k represent a positive integer, so that 2k represents a positive even integer, and suppose we want to answer the following question: How can 2k be given as a sum of two numbers x a nd y so that xy is as large as possible?

a) Answer the question specifically for 2k=10 by trying all possible pairs of integers whose sum is 10. Based on your answer, make a conjecture for the general case.

b. Prove the conjecture you made in part (a) by writing a quadratic function in factored form.

Answer 1

a) x^2 - (p+q+2k)x + (p+k)(q+k)

b) x^2 - (p+q-2k)x + (p-k)(q-k)

Answer 2

a)They are g(x) = (x - p)^2 + b(x - p) + c and h(x) = (x - q)^2 + b(x - q) + c
b) just change p to -p and q to -q for (b)

Second part is based on the fact that , given its perimeter, the rectangle of max area is a square. So K and K are the values. For 2k = 10, 10 and 10 are the parts as desired. the products of (1,19), (2, 18), (3,17),..... (9,11) will all give lesser product than (10,10) Check it.

This is the most general statement irrespective of whether quadratic is in factored form or not.

Answer 3

(x - p)(x - q) = x^2 - (p + q)x + pq = x^2 + bx + c

so p + q = -b and pq = c

(x - (p + k))(x - (q + k)) has zeros p + k and q + k

= x^2 - (p + k + q + k)x + (p + k)(q + k)
= x^2 - (p + q + 2k)x + pq + (p + q) k + k^2
= x^2 - (-b + 2k)x + (c + (-b)k + k^2)

or x^2 - (2k - b)x + (k^2 - bk + c) = 0 is the first equation

Answer 4

f(x) = x^2 + bx + c has zeros p and q.
thus f(x) = (x-p)(x-q) thus c = pq, b = -(p+q)

a) replace p with p+k, q with q +k
c=(p+k)(q+k) and b = -(p+q+2k)

the wanted f(x) = x^2 +-(p+_q+2k)x + (p+k)(q+k)

If you understand this you also can make theother questions