SOLUTION: Make a graph of y=-x^2-2x+3 and find the vertex, y-intercept, extra points. I don't understand how I can find the vertex, y-intercept and extra points.

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: Make a graph of y=-x^2-2x+3 and find the vertex, y-intercept, extra points. I don't understand how I can find the vertex, y-intercept and extra points.       Log On


   

Question 437257: Make a graph of y=-x^2-2x+3 and find the vertex, y-intercept, extra points.
I don't understand how I can find the vertex, y-intercept and extra points.
Found 2 solutions by Alan3354, Gogonati:
Answer by Alan3354(69443) About Me  (Show Source):
Answer by Gogonati(855) About Me  (Show Source):
You can put this solution on YOUR website!
graph%28300%2C+300%2C+-5%2C+5%2C+-5%2C+5%2C+-x%5E2-2x%2B3%29
This graph represent a parabola: y=-x%5E2-2x%2B3=-%28x%5E2%2B2x%2B1%29%2B1%2B3=-%28x%2B1%29%5E2%2B4
From the final expression we see that the vertex is: (-1, 4)
To find the y-intercept we make x=0: y=-0^2-2*0+3=3; (0, 3)
Other extra points are the x-intercepts: -x^2-2x+3=(x+3)(x-1)=0: x=-3 and x=1;
x-intercepts are:(-3, 0) and (1, 0).
Done.