How do you find the point of intersection between linear and quadratic relation: y=x+7 and y=(x+3)^2-8?

Jun 24, 2017

$\left(- 6 , 1\right) \text{ and } \left(1 , 8\right)$

Explanation:

$\text{equate " y=x+7" and } y = {\left(x + 3\right)}^{2} - 8$

$\Rightarrow {\left(x + 3\right)}^{2} - 8 = x + 7$

$\text{rearrange and equate to to zero}$

${x}^{2} + 6 x + 9 - 8 = x + 7$

$\Rightarrow {x}^{2} + 5 x - 6 = 0$

$\Rightarrow \left(x + 6\right) \left(x - 1\right) = 0$

$\Rightarrow x = - 6 \text{ or } x = 1$

$\text{substitute these values into } y = x + 7$

$x = - 6 \to y = - 6 + 7 = 1 \Rightarrow \left(- 6 , 1\right)$

$x = 1 \to y = 1 + 7 = 8 \Rightarrow \left(1 , 8\right)$

$\text{the points of intersection are " (-6,1)" and } \left(1 , 8\right)$ graph{(y-x^2-6x-1)(y-x-7)=0 [-20, 20, -10, 10]}