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The graph of a function is a line that passes through the points (3, 17) and (6, 32). What is the equation of this function?

 Jun 9, 2021
 #1
avatar+1006 
+1

My math is very rusty, admittedly, so perhaps I shouldn't pick back up the keyboard to help, but I still will regardless. If I recall, this is going to be a point-slope formula problem, followed by a slope-intercept formula problem.

 

 

Point slope formula: \(y − y_{1} = m(x − x_{1})\)

 

Substitute the values accordingly: Treat \((3, 17)\) as \(y\) and \((6, 32)\) as \(y_{1}\).

 

Post-substitution: \(17 - 32 = m(3 - 6)\)

 

From there, simplify as needed.

 

Step 1: \(17 - 32 = m(3 - 6)\)

 

Step 2: \(-15 = m(-3)\)

 

Step 3: \(\frac{-15}{-3} = \frac{m(-3)}{-3}\)

 

Step 4: \(5 = m\)

 

 

From there, take either point and, knowing the value of \(y\)\(m\), and \(x\), plug in those numbers to the slope intercept formula to find \(b\) and make sure they match.

 

Slope intercept formula: \(y=mx+b\)

 

Substitutions: \(17=5(3)+b\) and \(32=5(6)+b\)

 

Calculations 1: \(17=15+b\) and \(32=30+b\)

 

Calculations 2: \((17-15)=(15-15)+b\) and \((32-30)=(30-30)+b\)

 

Solution(s): \(2=b\) for both equations.

 

 

The likely solution is that your answer wants to be done up in slope-intercept form, in which case you would plug in your values for \(m\) and \(b\), which would yield a final answer of \(y=5x+2\). Hope this helped with understanding how you reached such an answer!

 Jun 9, 2021
 #2
avatar+114462 
+2

WELCOME BACK GOLDENLEAF!    cool

 

Your method looks fine. (I have not checked all your working)

 

 

I do this in a way that is not quite mathematically correct but it works fine.

any equation of a line question I start with 

 

gradient = gradient

\(\frac{rise}{run}=\frac{rise}{run}\\ \frac{y-y_1}{x-x_1}=\frac{y_2-y_1}{x_2-x_1}\\ \frac{y-17}{x-3}=\frac{32-17}{6-3}\\ \)

and then just simply

 

 

Technically this presentation is not quite right because it would make a hole at x=3, but just ignore that.

 Jun 9, 2021
edited by Melody  Jun 9, 2021
 #3
avatar+1006 
+1

Good to be back! Albeit only on rare occasion, since I'm not in school anymore (had to drop out due to finances). And I haven't taken a maths course (to completion) since... gosh, 2016? But I'll still be here to keep my mind sharp, and to stay learned!

GoldenLeaf  Jun 12, 2021

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