Question 586230
I do the calculation as {{{C=5(F-32)/9}}}, which is the same as {{{C=(5/9)(F-32)}}}.  That is the equation in point-slope form when I take freezing as my point. At the freezing point of pure water, C=0, F=32, the point (32,0). You could call it the F intercept. Your equation, {{{C=(5/9)F-160/9}}} is equivalent, but would be considered the slope-intercept form.
a) The slope is {{{5/9}}}, the coefficient of the independent variable F. The C intercept is the value of C when F=0, {{{-160/9}}}. The intercept point (0, {{{-160/9}}}.) is the point where the C versus F curve crosses the C axis. (We are using C where we usually use x, and F where we usually use y).
b) The slope is the temperature change in degrees Celsius for a change of 1 degree Fahrenheit. The C intercept is the Temperature in degrees Celsius, when the reading in degrees Fahrenheit is zero. It is approximately -18 (rounding to the nearest integer).
c) {{{C=(5/9)F-160/9}}} ---> {{{9C=5F-160}}} (multiplying both sides of the equal sign times 9)
{{{9C=5F-160}}} ---> {{{9C+160=5F}}} ((adding 160 to both sides)
{{{9C+160=5F}}} ---> {{{(9/5)C+32=F}}} or  {{{F=(9/5)C+32}}} (dividing both sides by 5)
Now the slope is {{{9/5}}}, representing is the temperature change in degrees Fahrenheit for a change of 1 degree Celsius. The F intercept is 32, the temperature in degrees Fahrenheitthat corresponds to zero Celsius.