Hi, i don't understand how to solve this quadratic function: y=-3x^2. I can't seem to know how to find the axis of symmetry, vertex, and how to plot it. I Know how to solve questions like this y-5=-(x+2)^2. But I don't know how to solve that basic function. Help me asap please. Thanks in advance.
2007-09-02 08:17:09 · 2 answers · asked by Here for You 2 in Science & Mathematics ➔ Mathematics
I still do not understand...
2007-09-02 08:48:55 · update #1
Oh now I understand ^^. Thank you very much ktm
2007-09-02 11:35:55 · update #2
y = -3x^2 is symmetric with respect to the y axis. The y axis is your line of symmetry. Try and graph y = -3x^ + 0x and see if that helps you run through your graphing steps! Good luck!
Additional:
If you need to graph this quad without aid of a calculator, your teacher probably told you to use the vertex formula to find both axis of symmetry and your max or min point,
v = -b/2a your quadratic has no b term (ax^2+bx+c)
so -0/2(-3) = 0 That's why your axis of symmetry is 0 (meaning the y axis).
Now if you plug that x=0 into y = -3x^2, y = -3(0)^2 then y also is 0
Your vertex point is (0,0) The graph opens downward b/c your quad is negative. Now just find two additional points on either side of the vertex and you will have your graph.
2007-09-02 08:23:23 · answer #1 · answered by ktm 3 · 2⤊ 0⤋
If you understand y - 5 = -(x + 2)^2, then you should understand (y - 0) = -3(x - 0)^2, which is your problem.
2007-09-02 09:30:08 · answer #2 · answered by Tony 7 · 1⤊ 0⤋