Why is $x^2+x$ the same as writing $(x+0.5)^2-0.25$?

Question

I find it extremely weird that a translation vertically by a variable should cause the graph to move both vertically and horizontally. Also, why 0.5 and 0.25?

Answer 1

Well, from $(a+b)^2=a^2+2ab+b^2$, you have with $a=x$ and $b=0.5$: $$\left(x+0.5\right)^2 = x^2+2 \cdot x \cdot 0.5 + 0.5^2 = x^2+x+ 0.25$$ so you also have: $$\left(x+0.5\right)^2-0.25=x^2+x$$

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As for some graphical interpretation: you probably know that adding a number $c$ to $f(x)$, so changing $f(x)$ to $f(x)+c$, causes the graph to shift $c$ units in the vertical direction; upwards for positive $c$ and downwards for negative $c$.

Adding the variable $x$ isn't the same as adding a constant $c$, but if you imagine fixing at a certain $x$, you locally shift $x$ units. But $x$ is a variable: the larger $x$ is, the more you move the graph!

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Approaching it from the other perspective: shifting $x^2$ horizontally over $a$ units and vertically over $b$ units, changes the function to: $$\left( x-a \right)^2 +b = x^2-2ax+a^2+b$$ From this you can easily see that if you want to know the effect of simply adding $x$ to $x^2$, you need $-2a=1$ and $a^2+b=0$ from which the vertical and horizontal translations in question follow.

Answer 2

we get by the binmoial formula: $$x^2+2x\cdot 0.5+0.25-0.25=x^2+x$$ the answer is yes! you can find the answer above!

Answer 3

hint

\begin{align}(x+0.5)^2&=\left(x+\frac 12\right)^2\\ &=x^2+2x\frac 12+\left(\frac 12\right)^2\\ &=x^2+x+\frac 14\\ &=x^2+x+0.25\end{align}

A horizontal translation $+0.5$ followed by a vertical translation $-0.25$, give the same graph.

Answer 4

Consider a more general case: ($a > 0$) \begin{align} a \, x^2 + b \, x &= (\sqrt{a})^2 x^2 + 2 \cdot \frac{b}{2 \, \sqrt{a}} \cdot \sqrt{a} x + \frac{b^2}{4 \, a} - \frac{b^2}{4 \, a} \\ &= \left(\sqrt{a} x + \frac{b}{2 \, \sqrt{a}}\right)^2 - \frac{b^2}{4 \, a}. \end{align}

In this example the values of $(a,b)$ are $(1, 1)$.

Answer 5

Okay so $0.5 = \frac{1}{2}$ and $0.25 = \frac{1}{4}$.

Let's start from $$\left ( x + \frac{1}{2} \right )^{2} - \frac{1}{4}$$

$$= \left (\frac{2x+1}{2} \right )^{2} - \frac{1}{4}$$ $$= \frac{(2x+1)^{2}}{4} - \frac{1}{4}$$ $$= \frac{(2x+1)^{2}-1}{4}$$ $$= \frac{4x^{2}+4x}{4}$$

Then factor out $4x$ in the numerator to obtain $$= \frac{4x(x+1)}{4}$$

Then you'll get $x^{2}+x$.

Answer 6

This is called Completing the square intuitively we want to transform the equation so it doesn't have a linear term (the term with $x$).

So we tweak the well known formula for $(x+y)^2=x^2+2xy+y^2$ to get $$(x+y)^2-y^2=x^2+2xy$$ Now since we have that $2xy=x$ we get that $y=\frac{1}{2}$ which gives the above formula. This is particularly useful for finding the vertex of parabola.

Answer 7

We can use the completing the square method on $x^2+x$ to achieve this result:

Given an equation of the form $x^2+ax+b$ we can write this as $$\left(x+\frac a2\right)^2-\left(\frac a2\right)^2+b$$

So, in this scenario, we have $a=1$ and $b=0$, so we have \begin{align}x^2+x&=\left(x+\frac 12\right)^2-\left(\frac 12\right)^2+0\\ &=(x+0.5)^2-\frac 14\\ &=(x+0.5)^2-0.25\end{align}

Answer 8

To address one point about this question, the expression "$x^2 + x$" does not mean "translation vertically by a variable". In fact, "translation vertically by a variable" doesn't really make much sense (although the answer of @StackTD makes a game attempt at putting sense on it).

This does raise the question: What does $x^2+x$ mean geometrically?

First draw the two graphs $y=x^2$ and $y=x$ on the same coordinate plane.

You can now obtain the graph of $y=x^2+x$ by a geometric process. For each value of $x$, look on the vertical line with that $x$-coordinate; mark off $y$-coordinates on that vertical line, making it a copy of the real number line; observe the two points where that vertical line hits the two graphs; and add those two points using the standard method for adding two points on the number line, namely translate the first point vertically using the second point as the amount of translation.