SOLUTION: I am trying to find the vertex and intercepts so that I can graph the parabols I have gotten this far and my mind has gone totally blank. the equation reads ----- y = x^2 +2x

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: I am trying to find the vertex and intercepts so that I can graph the parabols I have gotten this far and my mind has gone totally blank. the equation reads ----- y = x^2 +2x       Log On


   

Question 12488: I am trying to find the vertex and intercepts so that I can graph the parabols I have gotten this far and my mind has gone totally blank.

the equation reads ----- y = x^2 +2x -3
-b/2a = -2/3
when I insert in the equation I get 4/9 + 24/9 - 33/9
which gives me -5/9
now I am lost
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
You had the right idea for finding the vertex of the parabola but got the wrong answer.
y+=+x%5E2+%2B+2x+-+3 Here: a = 1, b = 2, and c = -3, so:
%28-b%29%2F2a+=+-2%2F2 = -1 So the x-coordinate of the vertex is -1. To find the y-coordinate, substitute -1 for x in the equation and solve for y.
y+=+%28-1%29%5E2+%2B+2%28-1%29+-+3
y+=+1+-+2+-+3
y+=+-4 The y-coordinate is -4.
The vertex is at (-1, -4)
The x-intercepts are found by setting y = 0 and solving for x (there will be two of them).
0+=+x%5E2+%2B+2x+-+3 Solve by factoring.
%28x+-+1%29%28x+%2B+3%29+=+0
x-1+=+0 and x+%2B+3+=+0
x+=+1 and x+=+-3
The y-intercept is found by setting x = 0 and solving for y.
y+=+0%5E2+%2B+2%280%29+-+3
y+=+-3
Let's see what the graph look like:
graph%28300%2C200%2C-6%2C6%2C-6%2C6%2Cx%5E2%2B2x-3%29