SOLUTION: Please I need some extra help with this problems!! Graphing Parabolas, for example this y= -x^2-2x+1? please explain me how you do it? Thanks, Brian

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: Please I need some extra help with this problems!! Graphing Parabolas, for example this y= -x^2-2x+1? please explain me how you do it? Thanks, Brian       Log On


   

Question 439903: Please I need some extra help with this problems!!
Graphing Parabolas, for example this y= -x^2-2x+1?
please explain me how you do it?
Thanks, Brian
Found 2 solutions by stanbon, MathLover1:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Graphing Parabolas, for example this y= -x^2-2x+1?
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If you don't know much about quadratics and parabolas
the best you can do is plot points.
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If you know somethings you could find the y-intercept (0,1);
you could find the x-intercepts by solving -x^2-2x+1 = 0;
you could find the vertex at (-b/(2a),f(-b/(2a));
you could notice the parabola opens downward because the
coefficient of x^2 is negative.
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graph%28400%2C300%2C-10%2C10%2C-10%2C10%2C-x%5E2-2x%2B1%29
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Cheers,
Stan H.

Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!
Graphing the parabola: +y+=+ax%5E2+%2B+bx+%2B+c
1. Determine whether the parabola opens upward or downward.
a. If a+%3E+0, it opens upward.
b. If a+%3C+0, it opens downward.
2. Determine the vertex.
a. The x-coordinate is -b%2F2a
b. The y-coordinate is found by substituting the x-coordinate in the equation +y+=+ax%5E2+%2B+bx+%2B+c.
3. Determine the y-intercept+by setting x+=+0.
4. Determine the x-intercepts (if any) by setting y+=+0,or solving the equation
+y+=+ax%5E2+%2B+bx+%2B+c.
5. Determine two or three other points if there are no x-intercepts

.example:
y=+-x%5E2-2x%2B1
1. a+%3C+0, it opens downward
2. the vertex.
a. The x-coordinate is -b%2F2a=-%28-2%29%2F2%28-1%29=2%2F-2=-1
b. substitute the x-coordinate in the equation y=+-x%5E2-2x%2B1
y=+-%28-1%29%5E2-2%28-1%29%2B1
y=+-1%2B2%2B1
y=+2
the vertex is at (-1,2)
3. Determine the y-intercept+by setting x+=+0.
y=+-%280%29%5E2-2%280%29%2B1
y=1
4. Determine the x-intercepts (if any) by setting y+=+0,or solving the equation
y=+-x%5E2-2x%2B1
0=+-x%5E2-2x%2B1.use quadratic formula
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
x+=+%28-%28-2%29+%2B-+sqrt%28%28-2%29%5E2-4%2A%28-1%29%2A1+%29%29%2F%282%2A%28-1%29%29+

x+=+%282+%2B-+sqrt%284%2B4+%29%29%2F-2%29+
x+=+%282+%2B-+sqrt%288%29%29%2F-2%29+
x+=+%282+%2B-+2.83%29%2F-2+
x+=+%282+%2B+2.83%29%2F-2+
x+=+4.83%2F-2+
x+=++-2.41+
or
x+=+%282+-+2.83%29%2F-2+
x+=+0.83%2F-2+
x+=++-0.41+
5. Determine two or three other points if there are no x-intercepts
since you have x-intercepts, you skip this step

let see the graph:
+graph%28+500%2C+500%2C+-10%2C+10%2C+-10%2C+10%2C-x%5E2-2x%2B1%29+