Question

5. What is the vertex in the equation x^2 – 6x + 4
explain

Answer 1

if the vertex is the point where the graph intersect the x axis, then set the equation equals to zero and solve for x

Answer 2

complete the square.

Answer 3

find the minimum of the function; that would be ur vertex, sorry dont set it to 0

Answer 4

no. complete the square.
do u know how?

Answer 5

take the first derivative... dont complete the sqaure

Answer 6

2x-6 =0 so x=3
plug 3 in the original eqtion, u get -5

(3,-5) is your vertex

Answer 7

ok.
u have
\[ \large y=x^2-6x+4 \]
\[ \large y-4+\color{red}{3^2}=x^2-6x+\color{red}{3^2} \]
\[ \large y+5=(x-3)^2 \]

Answer 8

so the vertex is (3,-5)

Answer 9

ok

Answer 10

http://www.algebra-class.com/vertex-formula.html

Answer 11

so if you have ax^2+bx+c ,based on the sign of a, the parabola opens up or down. if a is positive, the parabola opens up; if a is negative it opens down... so here a is 1>0 so it opens up

Answer 12

yea

Answer 13

thx

Answer 14

\[x1 = (-b+\sqrt{b^2 - 4*a*c})/(2*a); x2=(-b-\sqrt{b^2 - 4*a*c})/(2*a)\]

Answer 15

plugging in the values of both a, b and c, you will get your x1 and x2

Answer 16

good. now can you take it from there?

Answer 17

sorry but im kinda confused with the way you wrote your answer. hold a second

Answer 18

yea.

Answer 19

using a calculator unless you can simplify it which is what I am doing now

Answer 20

x1=(6+√6^2−(4∗1∗4))/(2∗1)
x2=(6−√6^2−(4∗1∗4))/(2∗1)

Answer 21

do you understand that?

Answer 22

thx