# Question #f69d9

Jul 4, 2017

$y = 3 x + 12$

#### Explanation:

This problem is asking us to find a line such that the line intersects the graph of $y = {x}^{3} + 4 {x}^{2} + 3 x + 12$ 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 $0$.

$y = {\left(0\right)}^{3} + 4 {\left(0\right)}^{2} + 3 \left(0\right) + 12$

$y = 12$

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

$\left(0\right) = {x}^{3} + 4 {x}^{2} + 3 x + 12$

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

$0 = \left({x}^{3} + 4 {x}^{2}\right) + \left(3 x + 12\right)$

$0 = {x}^{2} \left(x + 4\right) + 3 \left(x + 4\right)$

$0 = \left({x}^{2} + 3\right) \left(x + 4\right)$

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 $\left(- 4 , 0\right)$.

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

$m = \frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}} = \frac{12 - 0}{0 - \left(- 4\right)} = \frac{12}{4} = 3$

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

$y = 3 x + 12$