I want to shift a line best some points, regardless of the slope. For example, a vertical line will shift to the best by having actually its two y coordinates changed, <\$(x_1, y_1+some number)\$ \$(x_2, y_2+some number)\$>, similarly a horizontal line will be shifted under by having actually its x coordinates adjusted only. Any lines that space not horizontal or vertical, will also be change by having actually both of your x and also y coordinates changed. How have the right to this be done?  Decompose your "diagonal" shift into an horizontal shift and a upright shift. That does the job, doesn"t it? y=mx+b.

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Upward shift: y=mx+b+a (a=upward shift, an unfavorable = downward).

Right Shift: y=m(x-c)+b (c=right shift, an unfavorable = left shift).

Any Shift: y=m(x-c)+b+a. It looks choose that you are talking around linear functions. The equation is

\$y=mx+b\$

If you want to change the graph \$y_0\$ systems upwards and \$x_0\$ systems to the best then the equation becomes

\$y+y_0=m(x-x_0)+b\$

For downward shifting and shifting come the left you have actually to change the signs.

Numerical example

Suppose the initial role is \$y=2x+2\$. And now we want to transition it \$1\$ unit upward and also \$3\$ units to the right. The equation becomes

\$y+1=2(x-3)+2\$

Multiplying out the brackets

\$y+1=2x-6+2\$

Subtracting 1 on both sides

\$y=2x-6+2-1\$

\$y=2x-5\$

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edited may 28 "16 at 6:26
answered may 28 "16 in ~ 6:06
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For a much more general method to changing ANY function (line, parabola, sinusoidal, what have actually you), we can simply consider shifting any duty \$f(x)\$ horizontally or vertically.

\$\$f(x) = y\$\$

If we desire to transition the duty upwards through \$u\$, this is the very same as including \$u\$ to every \$y\$ worth the role produces. We can take into consideration a role \$f_2(x)\$ such the which is \$f(x)\$ shifted upwards through \$u\$.

If \$f(x) = y\$,

Then \$f_2(x) = y + u\$.

Subsequently, \$f_2(x) = f(x) + u\$

Now, let"s take into consideration horizontal move to the best by \$v\$. To attain this, we have actually to think about what it method to move the duty to the right. It way for every \$x\$, we"re walk to relocate its linked \$y\$ to \$x + v\$ instead. This is the very same as assigning \$x\$ the \$y\$ connected with \$x - v\$. Let"s build a function which does just that and also call the \$f_3(x)\$.

If \$f(x) = y\$,

Then \$f_3(x + v) = y\$.

Subsequently, \$f_3(x) = f(x - v)\$

Now, let"s speak we wanted to incorporate these concepts at the same time: we desire to relocate a duty upwards by \$u\$ and to the right by \$v\$ simultaneously.

\$\$f_4(x) = f(x - v) + u\$\$

Now, let"s apply this come lines.

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\$\$f(x) = mx + b\$\$

\$\$f(x - v) + u = mx - mv + b + u\$\$

The new equation for your line still has actually slope \$m\$, but you have a new \$y\$-intercept \$(-mv + b + u)\$.