Question

Determine the equation of the line that passes through the points of intersection of the graphs of the quadratic functions f(x) = x^2 – 4 and g(x) = – 3x^2 + 2x + 8.

Answer 1

Answer:
`x-2y-2=0`

Explanation:

Given :

`f(x) = x^2 - 4 \\ g(x) = - 3x^2 + 2x + 8`

Point of intersection :

`f(x)=g(x)\\x^2-4=-3x^2+2x+8\\4x^2-2x-12=0\\2x^2-x-6=0\\2x^2-4x+3x-6=0\\2x(x-2)+3(x-2)=0\\(x-2)(2x+3)=0\\x=2\,,\,(-3)/(2)`

`x=2\,;f(2)=2^2-4=0\\x=(-3)/(2)\,; f\left ( (3)/(2) \right )=\left ( (3)/(2) \right )^2-4=(-7)/(4)=-1.75`

So, we have points
`\left ( 2,0 \right )\,,\,\left ( -1.5,-1.75\ \right )`

Equation of line passing through two points
`\left ( x_1,y_1 \right )\,,\,\left ( x_2,y_2 \right )` is given by
`y-y_1=(y_2-y_1)/(x_2-x_1)\left ( x-x_1 \right )`

Let
`\left ( x_1,y_1 \right )=\left ( 2,0 \right )\,,\,\left ( x_2,y_2 \right )=\left ( -1.5,-1.75\ \right )`

So, equation is as follows :

`y-0=(-1.75-0)/(-1.5-2)\left ( x-2 \right )\\y=(-1.75)/(-3.5)\left ( x-2 \right )\\y=(1)/(2)(x-2)\\2y=x-2\\x-2y-2=0`