So I'm trying to get an intuition of why: \begin{align*} f(x) &= x^2, \\ g(x) &= x, \\ (f+g)(x) &= x^2 + x, \end{align*} simply moves a quadratics vertex to a new position, but otherwise is an identical graph.


Mathematically I can follow this, and understand that $$ x^2 + x = (x + 0.5)^2 - 0.25, $$ and so leads to a vertex of $(-0.5, 0.25)$. But when I try to think about this visually, I would expect the slope of $y = x$ on the negative axis to 'drag down' $x^2$, and on the positive side 'push up' $x^2$ and therefore get a graph that does not have an axis of symmetry.

For example $\sin(x) + x$: https://www.desmos.com/calculator/g8tku7mbzx

This graph with sin is visually what I would 'expect' to happen to the quadratic.

I can't intuitively see why the slope of $y = x$ doesn't modify the graph to do something different to what the math actually performs. And so struggle to have a mental model of what should happen.

Thanks in advance.

  • 2
    $\begingroup$ The "dragging down" and "pushing up" effects happen in such a way that the axis of symmetry shifts towards one side. $\endgroup$ – GoodDeeds Sep 18 '16 at 11:21
  • $\begingroup$ For large (positive or negative) values of $x$, the term $x^2$ dominates the sum $x^2+x$. This is very different from the $\sin(x)+x$ example. $\endgroup$ – Matt L. Sep 18 '16 at 11:50

As you wrote, $x^2+x=(x+1/2)^2-1/4$ so that this is just a translated version of the parabola.

If you think in terms of slopes, $(x^2)'=2x$ represents a straight line and $(x^2+x)'=2x+1$ too. You just change the intercept, hence adding $1$ has both the effect of a vertical or horizontal translation.

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