How do you write an equation of a line with slope = 2 and contains point (-1 , 4)?

The general format for the point-intercept quation is #y=mx+b#

Where #m=#slope and #b=#y-intercept

Given the slope is #2#, the equation now looks like #y=2x+b#

Given #P(-1,4)# we can solve for the #b# variable by substituting the #x# and #y# cordinates.

#4=2*(-1)+b#
#4=-2+b#
#b=6#

Therefore the equation that has the slope of #2# and passes through the point #(-1,4)# is #y=2x+6#

Here's a graphical look,

graph{2x+6 [-10, 10, -5, 10]}

We can use the point-slope formula to write an equation for this line. The point-slope formula states: #(y - color(red)(y_1)) = color(blue)(m)(x - color(red)(x_1))#

Where #color(blue)(m)# is the slope and #color(red)(((x_1, y_1)))# is a point the line passes through.

Substituting the slope and values from the point in the problem gives:

#(y - color(red)(4)) = color(blue)(2)(x - color(red)(-1))#

#(y - color(red)(4)) = color(blue)(2)(x + color(red)(1))#

We can solve this equation for #y# to put the equation in slope-intercept form. The slope-intercept form of a linear equation is: #y = color(red)(m)x + color(blue)(b)#

Where #color(red)(m)# is the slope and #color(blue)(b)# is the y-intercept value.

#y - color(red)(4) = (color(blue)(2) xx x) + (color(blue)(2) xx color(red)(1))#

#y - color(red)(4) = 2x + 2#

#y - color(red)(4) + 4 = 2x + 2 + 4#

#y - 0 = 2x + 6#

#y = color(red)(2)x + color(blue)(6)#