Tell the vertex and axis of symmetry then graph y=-x^2+3

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Mikeh

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Jul 25, 2006
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Tell the vertex and axis of symmetry. Then graph.

. . .y = -x^2 + 3

Will someone please help me with this one?
 
What's wrong with the answer you received in your other thread on this question? What have you done? Where are you stuck?

Please be specific. Thank you.

Eliz.
 
I have figured out that the vertex is (0,3) and the axis of symmetry is (0)
but now they will when I try to factor for x intercepts I can't figure it out.
I tried to use the quadratic formula and can't figure out where to plot my points..
 
X-intercepts are found by plugging "zero" in for y, and solving. This can be done with the Quadratic Formula, though that is probably over-kill in this case.

Graphing is done by picking a few x-values, plugging them into the equation to find the corresponding y-values, drawing the dots, and sketching in the curve.

What steps have you tried? Where are you stuck? Please be specific. Thank you.

Eliz.
 
I could not figure out where to get the x values to plug in to the equation.
But I think you helped Thank you...
 
Mikeh said:
I could not figure out where to get the x values to plug in to the equation.
Click to expand...
You pick any x-values you want.

Eliz.
 
No it still doesn't work, my parabala is up-side down.

I used -2,and got 7
-1, and got 4
0, and got 3
1, and got 4
2, and got 7
 
Mikeh said:
I used -2,and got 7...
Click to expand...
How did you get that -(-2)<sup>2</sup> + 3 = -4 + 3 equalled 7...?

Thank you.

Eliz.
 
I forgot to put the - in front but went back and did it and it finally worked.
Yeah I finally got it. Thank you soooo much for your help....
 
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