March 3, 2015

Homework Help: math help,algebra

Posted by jas20 on Tuesday, February 20, 2007 at 4:02pm.

Okay this is what i have to do but i think i am doing something wrong.

directions are: Identify the axis of symmetry, create a suitable table of values, and sketch the graph (including the axis of symmetry).

The problem is:

y = -x^2+6x-2

so i did my table and looks like this

x y
-1 -9
0 -2
1 3
2 6

I've plotted those points into a graph and i get a diagnol line not a parabola. Also for the axis of symmetry i know that i have to use the formula

x = (-b)/(2a)

i don't understand how to get the axis of symmetry.

Your equation: y = -x^2 + 6x - 2

If the coefficient of x^2 is less than 0, then the parabola opens down; if the coefficient of x^2 is greater than 0, then the parabola opens up.

Substitute values for x and solve for y to get a few points to graph. Be sure you graph enough points to form the parabola. You can start with the vertex, which will give you one point to graph. The vertex is -b/2a, (4ac - b^2)/4a for (x, y). The axis of symmetry is the x-value.

a = -1, b = 6, c = -2

To find the vertex:
-b/2a = -6/2*-1 = -6/-2 = 3 -->this is the x-value.
(4ac - b^2)/4a = [(4*-1*-2) - 6^2]/4*-1 = -28/-4 = 7 -->this is the y-value.

Vertex = (3, 7)

Axis of symmetry is x = 3.

I hope this will help you get started.

Yes, I thought i posted a message earlier that i actually figured it out don't know how but i did and what you said is what i got. Thank you very much although i do need help in another problem that i keep on trying but i can't get i will post it up.thanks

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