Question

Determine whether the graph of y = x2 − 6x + 3 has a maximum or minimum point, then find the maximum or minimum value.

Minimum; (-6, 3)

Maximum; (-6, 3)

Minimum; (3, -6)

Maximum; (3, -6)

Answer 1

it's a parabola,and it has a SQUARED "x" variable, meaning is moving vertically
the "a" component is positive, meaning is opening upwards

how do you find the vertex point?, well \(\bf \left(-\cfrac{b}{2a}, c-\cfrac{b^2}{4a}\right)\)

Answer 2

right ? \[\left(\begin{matrix}-6 \\ 2x\end{matrix}\right)\]\[3 -\frac{ 6^2 }{ 4x }\]

Answer 3

hmm \(\Large\begin{matrix}
y = &x^2& − 6x& + 3\\
y = &1x^2& − 6x& + 3\\
&a\ \ \ &b\ \ \ &c
\end{matrix}\)

Answer 4

so \(\bf \left(-\cfrac{(-6)}{2(1)}, 3-\cfrac{(-6)^2}{4(1)}\right)\)

Answer 5

ugh idk :(

Answer 6

well, what's \(\bf -\cfrac{(-6)}{2(1)}\)

Answer 7

or \(\bf \cfrac{-1(-6)}{2(1)}\) if you prefer :)

Answer 8

2

Answer 9

6/2 = 2?

so what's 4/2 then?

Answer 10

no 3

Answer 11

so x = 3
what about \(\bf \cfrac{(-6)^2}{4(1)}\)

Answer 12

6?

Answer 13

-6

Answer 14

\(\bf \cfrac{(-6)^2}{4(1)} \implies \cfrac{-6 \times -6}{4 \times 1}\)

Answer 15

36 over 4

Answer 16

yes, so if you simplify 36/4?

Answer 17

i got 9

Answer 18

ahhh 9, so \(\bf 3-\cfrac{(-6)^2}{4(1)} \implies 3-(9)\)

Answer 19

so the vertex is at x = 3, y = -6
(3, -6)

Answer 20

thanks

Answer 21

yw

Answer 22

minimum or maximum

Answer 23

it's a parabola,and it has a SQUARED "x" variable, meaning is moving vertically
the "a" component is positive, meaning is opening upwards

Answer 24

so what do you think? minimum or maximum?

Answer 25

minimum is how LOW a function gets in an interval
maximum is how HIGH it gets in an interval

so you can think of them this way
a minimum is a burrow
a maximum is a hump

Answer 26

minimum

Answer 27

:)

Answer 28

:) thank you

Answer 29

np