How do you graph #y=4x#?

Graph #y=4x#
First, determine your slope and y-intercept.
(Your equation is in slope-intercept form)
Slope-intercept Form: #y=mx+b#
#m# is the slope
#b# is the #y#-intercept

In your equation, the slope is #4# and the #y#-intercept is #0#.

Because your #y#-intercept is #0#, place your first point at #(0,0)#. Your second point should be at #(1,4)# because the graph rises #4# units for every #1# unit is runs.

graph{y=4x [-10, 10, -5, 5]}

Take for x any two numbers. Let the x be 1 and 2.
So, according to the equation the results for the y would be as follows:
# y=4*1=4 # so this is #(x,y)=(1,4), #and
#y=4*2=8 => (2,8)#
#
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The number in front of the #x# is the gradient (slope). This number represents the amount of up or down for a given amount of along. This is determined reading from left to right.

The value of 4 is really #4/1#. This means that for 1 along you go up 4.

If the number in front of the x (coefficient) is positive the 'slope' goes up. If the coefficient is negative it means the 'slope' is down.

For this question; substitute any value you chose into #x#, multiply it by 4 and you have your value for y
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Let #x=0 -> y=4(0)#

At #x=0 " we find that "y=0#

So this can be our first point on the graph.

#color(blue)("Point"_1 -> (x_1,y_1)->(0,0))#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Choosing a number at random: I chose 2

Let #x=2 ->y=4(2) #

At #x=2 " we find that "y=8#

So this can be our second point on the graph.

#color(blue)("Pont"_2->(x_2,y_2)->(2,8))#
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Note that gradient #->("change in y axis")/("change in x axis") -> (y_2 -y_1)/(x_2-x_1)#

#=(8-0)/(2-0)" " =" "8/2" "=" "4/1" "# reading left to right.
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Mark your two points on the graph paper. Draw a line through them extending it to the edges of the axis.

#color(red)("Label your axis "-> " extra marks")#
#color(red)("Label your graph" ->" extra marks")#

The label could read something like:

"graph of #y=4x#"