gp3-hw1 college algebra solutions
TRANSCRIPT
GP3-‐HW1 College Algebra
Solution
(a) We want to describe the x-‐coordinates of the points on the parabola that are below the x-‐axis
𝑥|𝑥 < −1 or 𝑥 > 4
(b) We want to describe the x-‐coordinates of the points on the parabola that are above the x-‐axis
𝑥|− 1 ≤ 𝑥 ≤ 4
Solution
(a) We want to describe the x-‐coordinates of the points on the graph of 𝑓 that are below the points on the graph of 𝑔; 𝑥|𝑥 < −3 or 𝑥 > 1
(b) We want to describe the x-‐coordinates of the points on the graph of 𝑓 that are above the points on the graph of 𝑔; 𝑥|− 3 ≤ 𝑥 ≤ 1
Solution The graph of 𝑓 𝑥 = 𝑥! + 3𝑥 − 10 is a parabola that opens up. We want to describe the x-‐coordinates of the points on the graph of 𝑓 that are above the x-‐axis. Find the x-‐intercepts: 0 = 𝑥! + 3𝑥 − 10 0 = 𝑥 + 5 𝑥 − 2 𝑥 = −5 or 𝑥 = 2
𝑥|𝑥 < −5 or 𝑥 > 2
Solution The graph of 𝑓 𝑥 = 𝑥! + 8𝑥 is a parabola that opens up. We want to describe the x-‐coordinates of the points on the graph of 𝑓 that are above the x-‐axis. Find the x-‐intercepts: 0 = 𝑥! + 8𝑥 0 = 𝑥 𝑥 + 8 𝑥 = 0 or 𝑥 = −8
𝑥|𝑥 < −8 or 𝑥 > 0
Solution The graph of 𝑓 𝑥 = 𝑥! − 1 is a parabola that opens up. We want to describe the x-‐coordinates of the points on the graph of 𝑓 that are below the x-‐axis. Find the x-‐intercepts: 0 = 𝑥! − 1 0 = 𝑥 + 1 𝑥 − 1 𝑥 = −1 or 𝑥 = 1
𝑥|− 1 < 𝑥 < 1
Solution 𝑥! + 7𝑥 + 12 < 0 The graph of 𝑓 𝑥 = 𝑥! + 7𝑥 + 12 is a parabola that opens up. We want to describe the x-‐coordinates of the points on the graph of 𝑓 that are below the x-‐axis. Find the x-‐intercepts: 0 = 𝑥! + 7𝑥 + 12 0 = 𝑥 + 4 𝑥 + 3 𝑥 = −4 or 𝑥 = −3
𝑥|− 4 < 𝑥 < −3
Solution 6𝑥! − 5𝑥 − 6 < 0 The graph of 𝑓 𝑥 = 6𝑥! − 5𝑥 − 6 is a parabola that opens up. We want to describe the x-‐coordinates of the points on the graph of 𝑓 that are below the x-‐axis. Find the x-‐intercepts: 2𝑥 − 3 3𝑥 + 2 < 0 2𝑥 − 3 = 0 or 3𝑥 + 2 = 0 2𝑥 = 3 3𝑥 = −2
𝑥 =32 𝑥 =
−23
𝑥|−23 < 𝑥 <
32
Solution The graph of 𝑓 𝑥 = 𝑥! + 2𝑥 + 4 is a parabola that opens up. We want to describe the x-‐coordinates of the points on the graph of 𝑓 that are above the x-‐axis. Find the x-‐intercepts: 𝑥! + 2𝑥 + 4 = 0 𝑎 = 1, 𝑏 = 2, 𝑐 = 4 𝑏! − 4𝑎𝑐 = 2! − 4 1 4 = −12, so the parabola does not cross the x-‐axis. There are no x-‐intercepts. Find the coordinates of the vertex;
ℎ =−𝑏2𝑎 =
−22 1 = −1
𝑘 = 𝑓 −1 = −1 ! + 2 −1 + 4 = 3 Vertex ℎ, 𝑘 = −1, 3
𝑥|𝑥 is any real number
Solution 25𝑥! − 40𝑥 + 16 < 0 The graph of 𝑓 𝑥 = 25𝑥! − 40𝑥 + 16 is a parabola that opens up. We want to describe the x-‐coordinates of the points on the graph of 𝑓 that are below the x-‐axis. Find the x-‐intercepts: 25𝑥! − 40𝑥 + 16 = 0 5𝑥 − 4 5𝑥 − 4 = 0 5𝑥 − 4 = 0 5𝑥 = 4
𝑥 =45
No solution.
Solution 4𝑥! − 6𝑥 > −9 4𝑥! − 6𝑥 + 9 > 0 The graph of 𝑓 𝑥 = 4𝑥! − 6𝑥 + 9 is a parabola that opens up. We want to describe the x-‐coordinates of the points on the graph of 𝑓 that are above the x-‐axis. Find the x-‐intercepts: 4𝑥! − 6𝑥 + 9 = 0 𝑎 = 4, 𝑏 = −6, 𝑐 = 9 𝑏! − 4𝑎𝑐 = −6 ! − 4 4 9 = 36− 144 = −108, so the parabola does not cross the x-‐axis. There are no x-‐intercepts. Find the coordinates of the vertex;
ℎ =−𝑏2𝑎 =
62 4 =
34 = .75
𝑘 = 𝑓34 = 4
34
!
− 634 + 9 =
274 = 6.75
Vertex ℎ, 𝑘 = . 75, 6.75
𝑥|𝑥 is any real number
Solution The domain of a function is the set of all real numbers excluding; -‐ any values of the variable that cause division by zero, and -‐ any values of the variable that cause the even root (square root) of a negative number 𝑥 − 3𝑥! ≥ 0 −3𝑥! + 𝑥 ≥ 0 3𝑥! − 𝑥 ≤ 0 The graph of 𝑓 𝑥 = 3𝑥! − 𝑥 is a parabola that opens up. We want to describe the x-‐coordinates of the points on the graph of 𝑓 that are below the x-‐axis. Find the x-‐intercepts: 𝑥 3𝑥 − 1 = 0 𝑥 = 0 or 3𝑥 − 1 = 0 3𝑥 = 1
𝑥 =13
Domain = 𝑥|0 ≤ 𝑥 ≤13