Factorise it to:
(x+3)(x-1), then x=1 and x=-3 [THIS IS WHERE IT CROSSES X AXIS]
And it crosses y axis at y=3. Plot this graph with the common shape of a quadratic and you're done

Reply 4

thiss1

Original post by ImNormal

Factorise it to:
(x+3)(x-1), then x=1 and x=-3 [THIS IS WHERE IT CROSSES X AXIS]
And it crosses y axis at y=3. Plot this graph with the common shape of a quadratic and you're done

thats what i thought but (x+3)(x-1) gives you x2+2x-3 but it has to be y=x2+2x+3

Reply 5

ImNormal

Original post by thiss1

thats what i thought but (x+3)(x-1) gives you x2+2x-3 but it has to be y=x2+2x+3

My bad. It doesn't even have any real roots, both of its solutions are in the complex plane.

Reply 6

gdunne42

Original post by thiss1

I HAVE BEEN TRYING TO DO THIS QUESTION BUT DONT KNOW HOW ?

PLEASE HELP ME.

SKETCH A GRAPH OF y=x2+2x+3

If you have completed the square on this quadratic you can use it to find the minimum point of the curve.

(edited 9 years ago)

Reply 7

Slowbro93

Original post by ImNormal

Factorise it to:
(x+3)(x-1), then x=1 and x=-3 [THIS IS WHERE IT CROSSES X AXIS]
And it crosses y axis at y=3. Plot this graph with the common shape of a quadratic and you're done

This isn't the question the OP posted :no:

OP, you'll see that you can't factorise your equation so I would recommend completing the square. Can you see what do to now?

Reply 8

7253567

Original post by thiss1

I HAVE BEEN TRYING TO DO THIS QUESTION BUT DONT KNOW HOW ?

PLEASE HELP ME.

SKETCH A GRAPH OF y=x2+2x+3

Calculate the discriminant. If it is less than 0 it is a parabola above the x axis and so on.

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Reply 9

thiss1

Original post by Slowbro93

This isn't the question the OP posted :no:

OP, you'll see that you can't factorise your equation so I would recommend completing the square. Can you see what do to now?

yes i tried completing the square but

i got

x=1+- square root of -2

i dont know how to go from there

Reply 10

thiss1

Original post by Trouty97

Calculate the discriminant. If it is less than 0 it is a parabola above the x axis and so on.

Posted from TSR Mobile

thanks for your reply

i did that but i dont know what to do after i complete the square

i got

x=1+- square root of -2

i dont know how to go from there

Reply 11

gdunne42

Original post by thiss1

thanks for your reply

i did that but i dont know what to do after i complete the square

i got

x=1+- square root of -2

i dont know how to go from there

E.g.
y=x^2 + 4x + 5
Complete the square
y=(x + 2)^2 + 1
The minimum value of (x + 2)^2 is 0 therefore
The minimum point on the curve is where x+2=0 I.e. x=-2 and y=1

Reply 12

maame b

Op find the minimum point as this a positive quadratic curve. This gives it a minimum of -1 , 2 . Because the curve has no real roots it doesn't cross either the x or y axis so just hangs.

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