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:$x-2y-2=0$

Step-by-step explanation:

Given :

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

Point of intersection :

$f(x)=g(x)\\x^2-4=-3x^2+2x+8\\4x^2-2x-12=0\\2x^2-x-6=0\\2x^2-4x+3x-6=0\\2x(x-2)+3(x-2)=0\\(x-2)(2x+3)=0\\x=2\,,\,\frac{-3}{2}$

$x=2\,;f(2)=2^2-4=0\\x=\frac{-3}{2}\,; f\left ( \frac{3}{2} \right )=\left ( \frac{3}{2} \right )^2-4=\frac{-7}{4}=-1.75$

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 $y-y_1=\frac{y_2-y_1}{x_2-x_1}\left ( x-x_1 \right )$

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 :

$y-0=\frac{-1.75-0}{-1.5-2}\left ( x-2 \right )\\y=\frac{-1.75}{-3.5}\left ( x-2 \right )\\y=\frac{1}{2}(x-2)\\2y=x-2\\x-2y-2=0$