Question #f69d9

This problem is asking us to find a line such that the line intersects the graph of #y = x^3 + 4x^2 + 3x + 12# at the x and y intercepts of that graph (which would then also be the intercepts of that line).

To do this, all we have to do is find the #x# and #y# intercepts of the graph, and then draw the line passing through those two intercepts to complete the system.

Let's first solve for the #y# intercept by setting x equal to #0#.

#y = (0)^3 + 4(0)^2 + 3(0) + 12#

#y = 12#

So the y intercept is #12#, or #(0, 12)#. Now, we set y equal to #0# to find the #x# intercept:

#(0) = x^3 + 4x^2 + 3x + 12#

We can group the terms in groups of two, and then factor and redistribute to get the following:

#0 = (x^3 + 4x^2) + (3x + 12)#

#0 = x^2(x + 4) + 3(x+4)#

#0 = (x^2+3)(x+4)#

Since these two terms multiplied together gives us zero, this means that either #x^2+3# is equal to zero, or #x+4# is equal to zero. However, #x^2+3# must always be greater than 0, so our solution must be when:

#x+4 = 0#

#x = -4#

So our #x# intercept is #-4#, or #(-4, 0)#.

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Now, we have our two intercepts: #(-4,0)# and #(0, 12)#. The line connecting these two points will be our solution. Let's find the slope of this line:

#m = (y_2-y_1)/(x_2-x_1) = (12 - 0)/(0 - (-4)) = 12/4 = 3#

So our slope is 3, and our #y# intercept is 12. Using slope-intercept form, this means that the equation of the line is:

#y = 3x+12#

Final Answer