Question

Please Help

which of the following is the minimum value of the equation y=2x^2+5

A.0
B.2
C.5
D.-5

Answer 1

when dealing with a quadratic, we know the min (when the x^2 terms Is positive) is given by
x = -b/(2a)
where a,b c are ax^2+bx+c

Answer 2

or take the derivative. what class is this for?

Answer 3

this is for alg.2

Answer 4

ok use my first reply

Answer 5

what does a=?
what does b = ?

Answer 6

a is 2 Im not sure about B.

Answer 7

b is gone, so b = 0

Answer 8

so min x = -b/(2a) = ?

Answer 9

so its zero because 2(2) is 4 / 0 is 0

Answer 10

correct, x =0
but we want the min VALUE, value=y
we know the min is AT x=0
so plug in x=0 to the original function and what do you get?

Answer 11

5?

Answer 12

correct

Answer 13

thank you

Answer 14

so if we ask WHERE is the min value, we want an x

if we ask WHAT Is the min value we want a y

Answer 15

Thank you can you also help me solve one more problem?

Answer 16

sure

Answer 17

Identify the axis of symmetry of the parabola.

A.x=5
B.x=4
C.x=6
D.X=3

Answer 18

wait, first off.
up there ^^^^ on the last question
you wrote x=0 because 2*2 = 4/0 =0

Answer 19

this is not why x=0
x = -b/(2a) = -0/(2*2) = -0/4 = 0/4 = 0

Answer 20

4/0 does not make sense

Answer 21

a = 2, b = 0
\[x=\frac{-b}{2a}= \frac{-0}{2*2}=\frac{0}{4}=0\]

Answer 22

understand?

Answer 23

yes

Answer 24

ok onto the next equation

Answer 25

question*

Answer 26

its the line down the middle
what x value is that?

Answer 27

down the middle of the porabolla

Answer 28

parabola*

Answer 29

4

Answer 30

yep

Answer 31

would that be the answer?

Answer 32

yeah

Answer 33

Thank you