Question 73691
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Find an equation for line with y-intercept 3 
that is perpendicular to the line 

{{{y=(2/3)x-4}}}

Compare that to the "slope y-intercept" form

{{{y=m*x+b}}}

and you see that the slope is {{{m=2/3}}} 

Now we use the rule:

If a line has slope {{{A/B}}} then 
1. A line parallel to it will ALSO have slope {{{A/B}}}
2. A line perpendicular to it will have slope {{{-B/A}}}
(Note: if the slope is a whole number and
not a fraction, write it over 1, so it will be
in the form {{{A/B}}})

A line perpendicular to a line with slope {{{2/3}}} will
therefore by the 2nd rule above have slope {{{-3/2}}}

Now all we have to do is use the slope y-intercept form

{{{y=m*x+b}}}

this time with {{{m=-3/2}}} and {{{b=3}}}, so we have

{{{y=(3/2)x+3}}}

If {{{f(x)=x^2 + 5}}}, Find {{{f(a+b)-f(a)}}}. 

First find {{{f(a+b)}}} by replacing {{{x}}} by {{{(a+b)}}}

{{{f(a+b)=(a+b)^2+5}}}

Now find {{{f(a)}}} by replacing {{{x}}} by {{{a}}}

{{{f(a)=a^2+5}}}

So {{{f(a+b)-f(a)}}} = {{{((a+b)^2+5)-(a^2+5)}}} = {{{(a+b)^2+5-a^2-5}}} =

{{{(a+b)^2-a^2}}} = {{{(a+b)(a+b)-a^2}}} = {{{a^2+ab+ab+b^2-a^2}}} =

{{{2ab+b^2}}}

Edwin</pre>