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

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- Jul 2nd 2007, 11:37 AMGAdamsGraph Problem
Could somone please provide some pointers on how to tackle these sorts of questions. I find them a bit scary.

- Jul 2nd 2007, 11:56 AMJhevon
For the first question.

the graph given is $\displaystyle y = 2x^2 + x - 4$

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

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

$\displaystyle 2x^2 + x - 4 = 0$

or in other words, where $\displaystyle 2x^2 + x = 4$

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 $\displaystyle 2x^2 - x - 5 = 0$

this is where $\displaystyle 2x^2 + x - 4 = 2x + 1$

So draw the line $\displaystyle y = 2x + 1$, the x-values of the points of intersection are your solutions - Jul 3rd 2007, 11:15 PMGAdams
- Jul 4th 2007, 11:47 AMJhevon
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 $\displaystyle 2x^2 - x - 5 = 0$, but i know nothing about that equation from the graph, all i know about from the graph is the equation $\displaystyle y = 2x^2 + x - 4$, 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:

$\displaystyle 2x^2 - x - 5 = 0$ (remember, my objective is to extrapolate $\displaystyle 2x^2 + x - 4 = \mbox { something }$ 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:

$\displaystyle 2x^2 + x - 5 = 2x$

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

$\displaystyle 2x^2 + x - 4 = 2x + 1$

I changed the graph of $\displaystyle 2x^2 - x - 5 = 0$ 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