Answer:
<em>Solution; ( 1, 0 ), and ( - 3, 4 )</em>
Step-by-step explanation:
See procedure below;
First let us make these two functions ( y = -x + 1
, y = x^2 + x - 2 ) equivalent;
- x + 1 = x^2 + x - 2,
-x - x + 1 + 2 - x^2 = 0,
- 2x + 3 - x^2 = 0,
x^2 + 2x - 3 = 0,
Now factor the simplified equation;
x^2 + 2x - 3 = 0,
( x^2 - x ) + ( 3x - 3 ) = 0,
x ( x - 1 ) + 3 ( x - 1 ) = 0,
( x - 1 )( x + 3 ) = 0,
And solve for x;
x - 1 = 0, and x + 3 = 0,
x = 1, and x = - 3
Now substitute this value of x into the two functions as to receive the y values for each x - value;
y = - ( 1 ) + 1, <em>y = 0 for x = 1</em>,
y = ( - 3 )^2 - 3 - 2, y = 9 - 3 - 2, <em>y = 4 for x = - 3</em>,
<em>Solution; ( 1, 0 ), and ( - 3, 4 )</em>
