finding x- and y-intercept: y = (2x + 1)^3 + 27

[gshock123]

I really need to find the x and y intercept for the following equation:

y = (2x + 1)^3 + 27

Thank you for your help

 

[stapel]

To find y-intercepts, plug "zero" in for "x", and solve.

To find x-intercepts, plug "zero" in for "y", and solve.

Eliz.

 

[gshock123]

I did that I just wanted to find out if my answers were correct, raising (2x+1) to the 3rd was a little tricky.
and the y i was unsure b/c mx+b it should be 27 but when i plug zero its 28, is it both?
Thanks so much

 

[skeeter]

when x = 0, (2x + 1)^3 + 27 = 28, the y-intercept

to find the x-intercept(s), first note that (2x + 1)^3 + 27 is the sum of two cubes which has a pattern of factoring ... a^3 + b^3 = (a + b)(a^2 - ab + b^2)

(2x + 1)^3 + 3^3 =

[(2x + 1) + 3][(2x + 1)^2 - 3(2x + 1) + 3^2] =

(2x + 4)(4x^2+4x+1 - 6x-6 + 9) =

(2x+4)(4x^2 - 2x + 4) =

4(x + 2)(2x^2 - x + 2)

set the last expression equal to zero ... x = -2 is an obvious root (y-intercept).

you'll have to use the quadratic formula to find the roots of the quadratic factor ...'
I'll leave that to you.