HELP. Express the function y = 2x^2 + 8x + 1 in vertex form.
y= blank (x + blank) - blank

Question

HELP. Express the function y = 2x^2 + 8x + 1 in vertex form. 
y= blank    (x + blank) - blank

johnweldon1993 · Answer

Still need help on this one @Power2Knowledge ?

anonymous · Answer

Yes sir

johnweldon1993 · Answer

Alright well lets start by grouping the 'x' together

$$\large (2x^2 + 8x) + 1$$ okay? now it looks like a number can be factored out of both those terms inside the parenthesis...which is...?

anonymous · Answer

4

johnweldon1993 · Answer

2*   since 2 can be taken out of 2 and 8

so we factor that out..

$$\large 2(x^2 + 4x) + 1$$

see what happened there?

johnweldon1993 · Answer

Now we just need to complete the square...

anonymous · Answer

4+4x+ 1

johnweldon1993 · Answer

Not quite...so to complete the square 

$$\large (x^2 + 4x)$$

we take half the coefficient of the 'x' which is 4 here....so half that would be 2...

and then we square that...so 2^2 = 4

$$\large 2(x^2 + 4x + 4) + 1$$

now we also need to multiply that 4 we just got...and the leading coefficient...of 2....so 4 times 2 = 8    and we subtract that from the equation

$$\large 2(x^2 + 4x + 4) + 1 - 8$$

Now we are all set to simplify...with me so far?

anonymous · Answer

yes

johnweldon1993 · Answer

Good...now lets simplify it a bit

$$\large 2(x^2 + 4x + 4) - 7$$

and we can now write the parenthesis in terms of a perfect square

$$\large 2(x + 2)^2 - 7$$

There is our vertex form

anonymous · Answer

Thank you  
so y = 2 
x = (x+2)^2
-7