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Draw the graphs of the functions and points of intersection for x^2 -2 and 2x^2 + 4x + 1

Let f(x) = x^2 - 2 and g(x) = 2x^2 + 4x + 1. Find the points of intersection and draw the graph.


The points of intersection are at,

    \[ x^2 - 2 = 2x^2 + 4x + 1 \ \implies \ x^2 + 4x + 3 = 0 \ \implies \ (x+3)(x+1) = 0. \]

Thus, they intersect at x=-3 and x=-1. These correspond to the points (-3, 7) and (-1,-1). The graph is below.

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3 comments

  1. Anonymous says:

    Hello! I believe the first intersection point is wrong, it should be (7, 7), or at least that’s what I get. I could be wrong too. Thanks for the solutions!

    • RoRi says:

      I think the points above are right. You can check by substituting in the values to see if the curves intersect there. So, for (7,7) we have

          \begin{align*}    f(7) &= 7^2 - 2 = 47 \\   g(7) &= 2*(7^2) + 4*7 + 1 = 98 + 28 + 1 = 127. \end{align*}

      So, the graphs of f(x) and g(x) don’t go through the point (7,7) (so definitely don’t intersect there).

      The points we get above are (-3,7) and (-1,-1). We can check those:

          \begin{align*}    f(-3) &= (-3)^2 - 2 = 9 - 2 = 7  \\    g(-3) &= 2*(-3)^2 + 4*(-3) + 1 = 18 - 12 + 1 = 7,  \end{align*}

      and

          \begin{align*}   f(-1) &= (-1)^2 - 2 = -1 \\  g(-1) &= 2*(-1)^2 + 4*(-1) + 1 = 2 - 4 + 1 = -1.  \end{align*}

      So, it checks out, the graphs of both functions go through the points (-3,7) and they also both go through (-1,-1). Unless I copied the problem from the book wrong or something. (Which has been known to happen, and I don’t have my copy with me.)

      • Anonymous says:

        Yes, you are correct, I found my mistake substituting the values as they are. Thanks for the explanation!

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