Question 413665
Find a linear function h such that h(-1)=-5 and h(7)=-6. What is h(3/2)?
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First you interpret it as this problems:

Find the equation of the line that passes through the two points
(-1,-5) and (7,-6).  

Use the slope formula: 


     y2 - y1      (-6) - (-5)      -6 + 5     -1 
m = ---------- = -------------- = -------- = ----- = {{{-1/8}}}
     x2 - x1        7 - (-1)        7 + 1      8

Then use the point-slope formula:

y - y1 = m(x - x1)

y - (-5) = {{{-1/8}}}(x - (-1) )

   y + 5 = {{{-1/8}}}(x + 1)

   y + 5 = {{{-1/8}}}x - {{{1/8}}}

       y = {{{-1/8}}}x - {{{1/8}}} - 5

       y = {{{-1/8}}}x - {{{1/8}}} - {{{40/8}}}

       y = {{{-1/8}}}x - {{{41/8}}}

Now replace y by h(x)
 
    h(x) = {{{-1/8}}}x - {{{41/8}}}

That's the first part.
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What is h({{{3/2}}})?
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Substitute {{{3/2}}} for x in

    h(x) = {{{-1/8}}}x - {{{41/8}}}

    h({{{3/2}}}) = {{{-1/8}}}{{{(3/2)}}} - {{{41/8}}}

    h({{{3/2}}}) = {{{-3/16}}} - {{{41/8}}} 

    h({{{3/2}}}) = {{{-3/16}}} - {{{82/16}}}

    h({{{3/2}}}) = {{{-85/16}}}  
  
Edwin</pre>