My Math Forum Axis of Symmetry

 Elementary Math Fractions, Percentages, Word Problems, Equations, Inequations, Factorization, Expansion

 May 21st, 2014, 02:25 AM #1 Newbie   Joined: Apr 2013 Posts: 23 Thanks: 0 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").
May 21st, 2014, 01:10 PM   #2
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 Originally Posted by chrie 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

 May 22nd, 2014, 01:48 AM #3 Senior Member     Joined: Nov 2013 From: Baku Posts: 502 Thanks: 56 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) Thanks from chrie
 May 22nd, 2014, 03:01 AM #4 Senior Member   Joined: Apr 2014 From: Glasgow Posts: 2,093 Thanks: 701 Math Focus: Physics, mathematical modelling, numerical and computational solutions 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). Thanks from chrie

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