Question

Draw the graph
y=3x^2+4x
Please, describe in detail

Answer 1

(x-k)^2 =c(y-a)

Answer 2

y -intercept

let x = 0

y = 3(0)^2 + 4(0)

therefore y-intercept = (0,0)

now solve for x-intercept

set y to 0

0 = 3x^2 + 4x

solve for x

0 = x(3x + 4)

so x = 0 and x = -4/3

does that give you any hints?

Answer 3

Yes, but where is the vertex of the parabola?
In (0;0) ?

Answer 4

well to get vrtex you have to change it to the form akash said

Answer 5

Take it to the completed square form

Answer 6

change your equation into this form..
(x-k)^2 =c(y-a)

Answer 7

@sauravshakya C=0 ?

Answer 8

3(x^2 +4/3 x) = y

Answer 9

(x+2/3)^2 = y/3 + 4/9

Answer 10

y=3x^2+4x
=3(x^2+4/3 x)
=3{x^2+2*x*2/3 + (2/3)^2 -(2/3)^2}
=3(x+2/3)^2-4/3

Answer 11

(x+2/3)^2 = 1/3( y + 4/3)

Answer 12

Now, Hint the vertex of a parabloa in the form y=a(x-h)^2+k is (h,k)

Answer 13

Does it help?

Answer 14

@sauravshakya I need the other way, because this way is not for school

Answer 15

U mean u want to use direct formula

Answer 16

What is direct formula?

Answer 17

If y=ax^2+bx+c then x-coordinate of the parabola is -b/2a

Answer 18

got it?

Answer 19

Yes, -2/3

Answer 20

YEP, thats x-coordinate and now find y-coordinate by substituting x=-2/3 in y=3x^2+4x

Answer 21

okay, i understood, Thank you!

Answer 22

WELCOMX