The x-intercepts are where y = 0. So, set x^2 +2x +1 to zero and solve:

x^2 +2x +1 = 0, factor this

(x+1)(x+1) = 0

x+1 = 0

x = -1 This is the only x intercept.

The graph is a parabola with vertex at (-1,0). After locating that point, plug in a few other values for x, such as -2, 0, which will give you y=1, and -3,1 which will give you y=4. You should graph those five points: (-1,0), (-2,1), (0,1), (-3,4), (1,4), and you’ll see the parabola forming.

It factors into (x + 9)(x – 5), so once you place it to 0 you get x = -9 or 5. So those are the two x-intercepts. The vertex is (horizontally) midway between -9 and 5, so that’s at x = -2. on account that f(-2) = -40 9, the vertex is the element (-2, -40 9). putting all that mutually provides you with an outstanding graph.

To find the x-intercept(s), let y = 0.

0 = x^2 + 2x + 1

Solve by factoring and you get:

0 = (x+1)(x+1)

0 = x + 1

-1 = x

So the graph crosses the x-axis at (-1,0)

ok. there can only be one x- intercept. there can be only one x and one y intercept. to find the x and y intercepts, but the equation in standard form (ax+by=c) by adding the x2+2x together to get 4x. you now have y=4x+1. then you need to move the x to the other side of the equation, sosubtract it from both sides. you now have -4x+y=1. now, to solve for x, cover up the y and you have -4x=1. solve for x by dividing both sides by -4, so you now have the x intercept as (-.1/4,0). do the same to get the y intercept. cover up the x and you have y=1, and so the y intercept is (0,1). email me if you have any questions.