How do you find the point of intersection between linear and quadratic relation: y=x+7 and y=(x+3)^2-8?
1 Answer
Jun 24, 2017
Explanation:
"equate " y=x+7" and " y=(x+3)^2-8
rArr(x+3)^2-8=x+7
"rearrange and equate to to zero"
x^2+6x+9-8=x+7
rArrx^2+5x-6=0
rArr(x+6)(x-1)=0
rArrx=-6" or " x=1
"substitute these values into " y=x+7
x=-6toy=-6+7=1rArr(-6,1)
x=1toy=1+7=8rArr(1,8)
"the points of intersection are " (-6,1)" and " (1,8) graph{(y-x^2-6x-1)(y-x-7)=0 [-20, 20, -10, 10]}
