Henrietta graphed the equation fx 7x 2 Lewis then graphed

Henrietta graphed the equation f(x) = 7x - 2. Lewis then graphed an equation that was a line 3 units above Henrietta\'s line. Which equation could have been the one Lewis graphed?

Select one:

a. f(x) = 4x - 2

b. f(x) = 7x + 1

c. f(x) = 7x - 5

d. f(x) = 10x - 2

Solution

Given f(x) = 7x - 2

Lewis drawn a line parallel to  f(x) = 7x - 2 above 3 units to it.

The slope of parallel lines are equal.

The given line is in the slope-intercept form y = mx + c

Slope of given line is \'7\'

y-intercept in the given line is \'-2\'.

3 units above this line, y-intercept is \'-2+3\' = \'1\'

Therefore, slope of required line is \'7\' and y-intercept is \'1\'.

Required line equation is f(x) = mx+c

f(x) = 7x+1

Lewis drawn a line  f(x) = 7x+1.

Therefore, Option \'b\' is correct.