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May 21st, 2014, 03:25 AM   #1
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Axis of Symmetry

I have to sketch a parabola y=-0,01x^2 -2x
I can calculate or know this things: parabola opens downward, graph is flat, graph goes through the origin, one of its intercepts is the origin.
y intercept (0,0) x1 intercept (-200,0) x2 intercept (0,0) Vertex (-100,100)

However I have now idea in which way I can "calculate" the coordinate (50,-125) (with the concept "axis of symmetry").
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May 21st, 2014, 02:10 PM   #2
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Quote:
Originally Posted by chrie View Post
I have to sketch a parabola y=-0,01x^2 -2x
I can calculate or know this things: parabola opens downward, graph is flat, graph goes through the origin, one of its intercepts is the origin.
y intercept (0,0) x1 intercept (-200,0) x2 intercept (0,0) Vertex (-100,100)

However I have now idea in which way I can "calculate" the coordinate (50,-125) (with the concept "axis of symmetry").

It is difficult to understand your description. You may need to try and explain in a simpler way and give more details.

In case it helps, I have made some videos on basic concepts of high school maths.

www.thephysicsnotes.com
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May 22nd, 2014, 02:48 AM   #3
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Math Focus: Geometry
$\displaystyle y=-0,01x^2 -2x $

$\displaystyle
\begin{array}{l|r}
x & y \\ \hline
0 & 0 \\ \hline
-200 & 0 \\ \hline
-100 & 100 \\ \hline
& \\
\end{array}
$

put numbers for x and find y, then xonnect points of (x,y)
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May 22nd, 2014, 04:01 AM   #4
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The intercepts and the vertex are all the information you need to sketch the quadratic. So you're done already!

The axis of symmetry is just a vertical line of symmetry that runs through the vertex.

The coordinate (50, -125) is an arbitrary coordinate that has nothing to do with the axis of symmetry. If you know that x = 50, substitute it in to get the y-coordinate. If you know the y-coordinate, set y equal to it and solve for x. You'll get two answers, so one of them will be (50,-125) and the other will be (-250,-125). You can just solve the quadratic with y = -125 or you can try the following:

If you already know one solution, you can get the other if you know the x-coordinate of the vertex. Let's say I know (50,-125) is a coordinate on the curve and I want the other point (a,-125). I find the horizontal distance between the coordinate I know (50) and the vertex (-100). This distance (150) is then transferred over to the other side of the vertex (-100 - 150 = -250). Therefore, the other coordinate is (-250,-125).

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