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=-(1)/(3)+(1)/(3)\cdot\sqrt[]{19},`

and

`x=-(1)/(3)-(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(-(1)/(3)+(1)/(3)\cdot\sqrt[]{19})-4=1-\sqrt[]{19}-4=-3-\sqrt[]{19},`

on the other:

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

So the points of intersection are

`(-(1)/(3)+(1)/(3)\cdot\sqrt[]{19},-3-\sqrt[]{19}),`

and

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

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