Question
Graph each equation and find the point(s) of intersection, if any.The circle $(x-1)^{2}+(y+2)^{2}=4$ andthe parabola $y^{2}+4 y-x+1=0$
Step 1
The equation of the parabola is $y^{2}+4 y-x+1=0$. We can rewrite this equation in the form $y^{2}+4 y = x-1$. This is a parabola that opens to the right if $y>0$ and to the left if $y<0$. The vertex of the parabola is at $(1,-2)$. Show more…
Show all steps
Your feedback will help us improve your experience
Brittany Scott and 87 other Algebra educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
Graph each equation and find the point(s) of intersection, if any. $y=\frac{4}{x+2}$ and the circle $x^{2}+4 x+y^{2}-4=0$
Systems of Equations and Inequalities
Systems of Nonlinear Equations
Graph each equation and find the point(s) of intersection, if any. The circle $(x+2)^{2}+(y-1)^{2}=4$ and the parabola $y^{2}-2 y-x-5=0$
Transcript
Watch the video solution with this free unlock.
EMAIL
PASSWORD