Question

find vertex of y=3(x-3)^2+4

Answer 1

since it is already in vertex form ...
y = a (x-Vx) + Vy

where the vertex is (Vx, Vy)

Answer 2

so it would be (-3,-4) x=3

Answer 3

i think it should has power 2 for in bracket, right @amistre64 ?

Answer 4

\[y=3(x-3)^{2}+4\]

Answer 5

y=a(x+b) + c
your turning point will simply be (-b,c)
the b value in your example is -3 so it will be positive 3, just remember it's the opposite.

Answer 6

and the y value you should be able to see from looking at it.

Answer 7

so it would be (3,4) x=-3

Answer 8

?

Answer 9

just (3,4) not x=-3

Answer 10

vertex/turning point is at (3,4)

Answer 11

o sorry i need axis of symmetry too

Answer 12

oh k.
For the y-intercept, make x=0 in the equation
for the x-intercept, make y=0 in the equation
then solve the equation.

Answer 13

o 3

Answer 14

sorry, I do not understand what you just typed.

Answer 15

i meant x=3 lol sorry

Answer 16

let me solve it now, I'm pretty sure that is incorrect though, there should be two answers, give me a second.

Answer 17

so you want to find the x-intercept I believe and to do so we have to make y=0
y=3(x-3)^2 + 4
0=3(x-3)^2 + 4
3(x^2-6x+9) + 4 = 0
3x^2 -18x + 27 + 4 = 0
3x^2 -18x + 31 = 0
here you can use the quadratic formula where a=3, b=-18 and c=31
\[ x = \frac{-b \pm \sqrt{b2-4ac}}{2a}\]

Answer 18

and if you do that, you actually get no answer because there is in fact no x-intercept.

Answer 19

you can try plot the graph on your calculator and you will see that the turning point as found before is at (3,4) and so that is the lowest point and it doesn't go past there, so clearly there is no x-intercept.

Answer 20

got it thanks

Answer 21

no problem, medal thanks :)