Question

Calculate the points of intersection for the line y = -3x - 4 and the curvey = -3x2- 5x + 2.

Answer 1

Given the line

$y=-3x-4,$

and the curve

$y=-3x^2-5x+2,$

in order to find the point at which they intersect, we must consider the equation:

$-3x-4=-3x^2-5x+2.$

Since this is a quadratic equation, let's move everything to the left side so that it equals 0. In other words, let's add 3x², 5x and subtract 2 from both sides:

$-3x-4+3x^2+5x-2=0,$$3x^2+2x-6=0.$

Let's solve this equation using the general formula for quadratic equations:

$x=\frac{-2\pm\sqrt[]{4-4(3)(-6)}}{2(3)},$$x=\frac{-2\pm\sqrt[]{4+72}}{6},$$x=\frac{-2\pm\sqrt[]{76}}{6},$$x=\frac{-2\pm2\cdot\sqrt[]{19}}{6}\text{.}$

This gives us the following two values of x:

$x=-\frac{1}{3}+\frac{1}{3}\cdot\sqrt[]{19},$

and

$x=-\frac{1}{3}-\frac{1}{3}\cdot\sqrt[]{19}.$

Now we know the x-coordinate of the points of intersection. In order to get the y-coordinate, we substitute these values on either of the equations we were given to begin with. We'll do it on the line since it's easier:

On one hand:

$y=-3(-\frac{1}{3}+\frac{1}{3}\cdot\sqrt[]{19})-4=1-\sqrt[]{19}-4=-3-\sqrt[]{19},$

on the other:

$y=-3(-\frac{1}{3}-\frac{1}{3}\cdot\sqrt[]{19})-4=1+\sqrt[]{19}-4=-3+\sqrt[]{19}.$

So the points of intersection are

$(-\frac{1}{3}+\frac{1}{3}\cdot\sqrt[]{19},-3-\sqrt[]{19}),$

and

$(-\frac{1}{3}-\frac{1}{3}\cdot\sqrt[]{19},-3+\sqrt[]{19}).$

In the image, we can see the approximate values of these coordinates.