Sketch the curve y=4-(x+3)^2, showing the points where the curve crosses the x-axis and any minimum or maximum points.

This equation rearranges to give -y=(x+3)^2-4, which is very similar to our curve y=(x+3)^2-4 from before. In fact, replacing y with -y in an equation is equivalent to reflecting the curve through the x-axis. We then take the points (-5,0), (-1,0) and (-3,-4) from before and replace y with -y, giving (-5,0), (-1,0) and (-3,4). We have found where the new curve crosses the x-axis and its minimum/maximum. The graph is an inverted u-shape since we have a -x^2 in the equation so (-3,4) is a maximum point.

JI
Answered by Jonny I. Maths tutor

3921 Views

See similar Maths GCSE tutors

Related Maths GCSE answers

All answers ▸

How do i solve quadratic equations?


Where do the lines 2y = 4x + 2 and - 3x + y = 4 intersect?


Calculate the length of the side of the Triangle marked by x.


Solve 2(x+4)=4x-6 for x


We're here to help

contact us iconContact ustelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

MyTutor is part of the IXL family of brands:

© 2026 by IXL Learning