X and y intercepts

[jgarcia]

3x2+6x+1

Here I need to find then x and y intercepts again.

I am completely at a loss how to do this effectively.

y-int= 3x²+6x+1

3(0)²+6(0)+1=y


1=y

x-int

3x²+6x+1=0
3(x²+2x)=-1
3((x)(x+2))=-1
3((x)(x+2))+1=0


I have no clue, I can't factor this out

 

[HallsofIvy]

3x2+6x+1

Here I need to find then x and y intercepts again.

I am completely at a loss how to do this effectively.

y-int= 3x²+6x+1

3(0)²+6(0)+1=y


1=y

So you have found the y-intercept! Good!

x-int

3x²+6x+1=0
3(x²+2x)=-1
3((x)(x+2))=-1
3((x)(x+2))+1=0


I have no clue, I can't factor this out

You are missing the point of factoring. The basic rule is that "if ab= 0 then either a=0 or b=0 or both". If you do not have "= 0" there is no use in factoring. And, in fact, most quadratics cannot be factored using integer coefficients. Use "completing the square" or the quadratic formula.

From \(\displaystyle 3(x^2+ 2x)= -1\), \(\displaystyle x^2+ 2x= -1/3\). You should know that \(\displaystyle (x+ 1)^2= x^2+ 2x+ 1\) so we can "complete the square" by adding 1 to both side of the equation: \(\displaystyle x^2+ 2x+ 1=-1/3+1\), \(\displaystyle (x+ 1)^2= 2/3\). Take the square root of both sides and solve for x.