Could somone please provide some pointers on how to tackle these sorts of questions. I find them a bit scary.

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- July 2nd 2007, 12:37 PMGAdamsGraph Problem
Could somone please provide some pointers on how to tackle these sorts of questions. I find them a bit scary.

- July 2nd 2007, 12:56 PMJhevon
For the first question.

the graph given is

the x-intercepts give the x-values for which y = 0

so for this graph, the x-intercepts are the x-values such that:

or in other words, where

so the solutions to the first question is simply the x-intercepts, which you can tell by looking on the graph

For the second question:

we want the solutions to

this is where

So draw the line , the x-values of the points of intersection are your solutions - July 4th 2007, 12:15 AMGAdams
- July 4th 2007, 12:47 PMJhevon
we find solutions from graphs based on intersections. therefore we need to look for those desired intersections, or sometimes create them.

in the following problem we were asked to draw a line to find solutions to the equation , but i know nothing about that equation from the graph, all i know about from the graph is the equation , so i have to somehow create this graph from the new graph given so i can get the required intersections.

so we begin with the new graph given:

(remember, my objective is to extrapolate from this graph

since i need a +x i have to add 2 x's to -x, but since this is an equation, i have to add it to both sides, so i get:

now i need to get a -4 instead of a -5, so i add 1 to both sides, we get:

I changed the graph of to get this, so my solutions to the new equation are exactly the same solutions to the one asked. since this new equation has the graph that i have originally, i simply need to find the intersection points between my old graph and the line 2x + 1