Derive Slope-Intercept Formula

harpazo

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How do we derive the slope-intercept formula in the picture?

sDraw_2018-08-11_12-26-56.png
 

MarkFL

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The point-slope formula may be written as:

\(\displaystyle y=mx+y_1-mx_1\)

We see that when \(x=0\), we have:

\(\displaystyle y=y_1-mx_1\)

And so we know the \(y\)-intercept is \(\left(0,y_1-mx_1\right)\)

If we define:

\(\displaystyle b=y_1-mx_1\)

Then, the \(y\)-intercept is at \((0,b)\), and the line can be written as:

\(\displaystyle y=mx+b\)

Or, we could begin by saying the \(y\)-intercept is at the point on the \(y\)-axis \((0,b)\) and so a line having slope \(m\), will be described by (using the point-slope formula):

\(\displaystyle y-b=m(x-0)\)

Or:

\(\displaystyle y=mx+b\)
 
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harpazo

Pure Mathematics
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The point-slope formula may be written as:

\(\displaystyle y=mx+y_1-mx_1\)

We see that when \(x=0\), we have:

\(\displaystyle y=y_1-mx_1\)

And so we know the \(y\)-intercept is \(\left(0,y_1-mx_1\right)\)

If we define:

\(\displaystyle b=y_1-mx_1\)

Then, the \(y\)-intercept is at \((0,b)\), and the line can be written as:

\(\displaystyle y=mx+b\)

Or, we could begin by saying the \(y\)-intercept is at the point on the \(y\)-axis \((0,b)\) and so a line having slope \(m\), will be described by (using the point-slope formula):

\(\displaystyle y-b=m(x-0)\)

Or:

\(\displaystyle y=mx+b\)
How important is this topic in calculus 1?
 

MarkFL

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Being able to do simple analytic geometry is extremely important in the study of calculus.
 

harpazo

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Being able to do simple analytic geometry is extremely important in the study of calculus.
There's a chapter in Sullivan's College Algebra dedicated to Conics.
 

MarkFL

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Yes, the study of conic sections is revisited in pre-calc and calc II.