FURTHER MATHS QUESTION

[AISHAB]

Hi,

How do you find the equation of a line, from a graph?

Hefferman 2011 Q8

The graph does not begin at 0

[TheEagle]

Hi,

How do you find the equation of a line, from a graph?

Hefferman 2011 Q8

The graph does not begin at 0

Hey!

If you look carefully, the point where the line crosses the y axis is not really the y intercept because the corresponding x value isn't zero. Instead it is (75, 68) and if you choose another point, i.e: (85,60), then if you find the gradient by doing (y2-y1) / (x2-x1), which would give -0.8. So at the moment our equation is Y=-0.8x+c      Now let's substitute one of our points that we had chosen before to find c --> 68= -0.8 *75 + c --> c= 128       hence our equation is Y=-0.8x + 128 --> Closest is OPTION E!

Another way to do this question faster (only works on exam 1 due to giving us the options):
substitute a pair of points that are clear into all options and see which gives the correct option
i.e: (subbing point (75, 68))
Option A) 68 = 49- 1.09 * 75 --> 68 is not equal to -32.75
Option B) 68= 68+1.09 *75 --> 68 is not equal to 49.75
Option C) 68=68-0.76*75 --> 68 is not equal to11
Option D) 68= 68+0.76 *75 --> 68 is not equal to 125
Option E) 68= 125 - 0.76 *75 --> 68 is equal to 68 (CORRECT)