Question

I will give an medal or possibly 2 for answering these questions

[9.01] Determine whether the graph of y = x2 − 4x + 1 has a maximum or minimum point, then find the maximum or minimum value.

Minimum; (2, -3)
Maximum; (2, -3)
Minimum; (-3, 2)
Maximum; (-3, 2)
7. [9.02] What are the x-intercept(s) of the graph of y − 8 = x2 − 6x?

(−2, 0) and (4, 0)
(2, 0) and (−4, 0)
(2, 0) and (4, 0)
(−2, 0) and (−4, 0)
8. [9.02] What are the solutions of 5x2 = −15x?

x = 5, x = 3
x = 5, x = −3
x = 0, x = 3
x = 0, x = −3
9. [9.02] Where does the graph of y = 2x2 − 5x + 3 cross the x-axis?

Answer 1

[9.02] Where does the graph of y = 2x2 − 5x + 3 cross the x-axis?

(−1 over 2, 0) and (−3, 0)
(1 over 2, 0) and (3, 0)
(3 over 2, 0) and (1, 0)
(−3 over 2, 0) and (−1, 0)
10. [9.02] Solve x2 − 3x = 28.

x = −2, x = 14
x = 2, x = −14
x = 4, x = −7
x = −4, x = 7

Answer 2

satellite how did i know you would be back lol

Answer 3

just unlucky i guess

Answer 4

\[y = x^2 − 4x + 1\] leading coefficient is positive, so it opens up and has a minimum

Answer 5

ok then but if i did my math right should it be minumum (2,-3)?

Answer 6

first coordinate of the vertex is \(-\frac{b}{2a}\) which in your case is 2

Answer 7

second is what you get, which is suppose is -3
but the question as stated is in fact incorrect

Answer 8

so wait now im confused is minimum (2,-3) not right then?

Answer 9

that is what you are supposed to say, yes

Answer 10

oh ok then what about the rest

Answer 11

what is incorrect about the question?

Answer 12

but the minimum value is a number, it is \(-3\)
the ordered pair is the vertex

Answer 13

minimum point is what it says

Answer 14

if i ask you what the least amount of money you would pay me, you would give a number as an answer, not an ordered pair
this question was written by someone who is guilty of causing you confusion later on
no matter, just give them the answer they want

Answer 15

there is no such thing as a "minimum point"

Answer 16

Well im about done with this algebra thats why im trying to get help with these cause my final will be coming soon

Answer 17

good

Answer 18

well not soon but in december

Answer 19

\[ y − 8 = x^2 − 6x\] is the same as
\[y=x^2-6x+8\] the \(x\) intercepts you find by setting
\[x^2-6x+8=0\] and solving for \(x\)

Answer 20

factor as \((x-2)(x-4)=0\) and the it should be easy to solve

Answer 21

yea ok i got 7 now but what about the rest of the questions?

Answer 22

\\[5x^2 = −15x\] is the same as
\[x^2=-3x\] rewrite as \(x^2+3x=0\) and the factor as \(x(x+3)=0\) and solve that one

Answer 23

then if its 0, then i would think it should be -3? am i right

Answer 24

yes

Answer 25

woohoo im learning

Answer 26

two solutions \(x=0\) or \(x=-3\)

Answer 27

ok, but 9 really confuses me

Answer 28

\[y = 2x^2 − 5x + 3\] again set
\[2x^2-5x+3=0\] and solve by factoring

Answer 29

try
\[(x-1)(2x-3)=0\]

Answer 30

so, if kinda confused but im thinking it might be something like (3,0) for one half of the answer

Answer 31

no not \((3,0)\) solve the one i wrote above

Answer 32

oh ok, sure
like i said confusing i dont even know basically where to start but if i did it right would it be something like 3x-4 = 0? or am i not even close

Answer 33

then like 3x = 4
then divide into a fraction?

Answer 34

do you know how to solve
\[(x-1)(2x-3)=0\] for \(x\) ?

Answer 35

not exactly

Answer 36

set \(x-1=0\) first

Answer 37

ok then what

Answer 38

just add the one to the other side?

Answer 39

if \(x-1=0\) then \(x=1\) right

Answer 40

ok then what did i do wrong then in the equation you gave me

Answer 41

and also set \(2x-3=0\) and solve

Answer 42

2x = 3
then it would work out into to be 3 over 2 right?

Answer 43

yes

Answer 44

ok so the first half is (3/2, 0)

Answer 45

ok i got it now i got 9 for the final answer to be (3/2,0) and (1,0)

Answer 46

I only need 10 now