Question

Standard form of a quadratic function.
y=a(x-h) square +k

1. y=1/2(x+3) square +2
how would you graph this ?

Answer 1

@Hero

Answer 2

@jim_thompson5910

Answer 3

Start with the graph of y = x^2

Compress it vertically by a factor of 2 to get y = 1/2x^2

Then shift it over 3 units to the left to get y = 1/2(x+3)^2

Finally, shift the entire graph up 2 units to get y = 1/2(x+3)^2 + 2

Answer 4

Alternatively, you can plug in values of x to get corresponding values of y.

This will give you a series of points which you can then plot and draw a curve through to graph y = 1/2(x+3)^2 + 2

Answer 5

wait i dont get it really can you show by doing it like f(-2)=1/2(-2+3) square+2 and then try to work it out and i choose -2 because after i plotted -3 i looked and i saw i could use -2 or -4 but i dont know how to work it out

Answer 6

y = 1/2(x+3)^2 + 2

y = 1/2(-1+3)^2 + 2 .. Plug in x = -1

y = 4

So when x = -1, y is y = 4


So one point is (-1, 4)

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y = 1/2(x+3)^2 + 2

y = 1/2(0+3)^2 + 2 .. Plug in x = 0

y = 6.5

So when x = 0, y is y = 6.5

And another one point is (0, 6.5)

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y = 1/2(x+3)^2 + 2

y = 1/2(1+3)^2 + 2 .. Plug in x = 1

y = 10

So when x = 1, y is y = 10

And another one point is (1, 10)

-----------------------------------

Keep going to generate more points

Answer 7

i still dont get it but can you help me with this one as well, y=2x square

Answer 8

Do the same: plug in values of x to get corresponding values of y to get a series of points.

Answer 9

Plot those points and draw a curve through them.

Answer 10

ok so what would be the vertex and the axis of symmetry for it

Answer 11

y = 2x^2

is the same as

y = 2x^2 + 0x + 0

compare this to

y = ax^2 + bx + c

What are 'a', b and c?

Answer 12

a= 2 b=0 c=0

Answer 13

now plug them into

x = -b/(2a)

Answer 14

to find the x coordinate of the vertex and the axis of symmetry

Answer 15

oh no see thats a different one from what am doing the formula for this one is y=a(x-h) square +k

Answer 16

and they just gave me y=2x square which looks confusing

Answer 17

yeah the different forms can be confusing at first, but you'll get the hang of them

Answer 18

yea but how would i do that though?

Answer 19

do what, find the vertex?

Answer 20

yea and the axis of symmetry

Answer 21

a = 2, b = 0

plug them into

x = -b/(2a)

Answer 22

would that be zero

Answer 23

yes, so the axis of symmetry is x = 0

Answer 24

the x coordinate of the vertex is x = 0

Answer 25

oh ok thank you, your so my hero :D

Answer 26

you're welcome

Answer 27

keep in mind you still need the y coordinate of the vertex to get an ordered pair for the vertex