Question 160269
i think your problem is you started with an x value that is way out there (1927).
you are so far from the y intercept and the slope is so steep going down from left to right (or going up from right to left), that the y intercept will be very large because when x = 0, this represents the year 0.
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if you pose the problem differently, then the y intercept will be the value of y in the year 1927.
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you do that by making x = the year - 1927.
then for 1927, the value of x is 0, and for 1997, the value for x is 70.
then your graph will look much better.
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now you have 
x1 = 0, y1 = 33.5
x2 = 70, y2 = 3.3
your slope was correct and is (-3.02/7)
equation for slope intercept form of the equation of the line is {{{y = m*(x-x1) + y1}}}
this becomes y = (-3.02/7)*x - (-3.02/7)*0 + 33.5
this becomes y = (-3.02/7)*x + (3.02/7)*0 + 33.5
this becomes y = (-3.02/7)*x + 33.5
your y-intercept is 33.5 as it should be since this is when x equals 0 and when x = 0, the year is 1927.
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in your original equation, x1 was 1927 and y1 was 33.5
slope was the same (-3.02/7)
slope intercept form of the equation should have been y = (-3.02/7)*(x - 1927) + 33.5 which becomes
y = (-3.02/7)*x - (-3.02/7)*1927 + 33.5
which becomes
y = (-3.02/7)*x + (3.02/7)*1927 + 33.5
which becomes
y = (-3.02/7)*x + 831.26285... + 33.5
which becomes
y = (-3.02/7)*x + 864.8628...
which becomes 864.9
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everything you did was correct, and if you let the x values remain as you originally made them, the y-intercept should be 864.9 because that is the year 0.