Question 3854
the y-intercept is where x=0, we then get y=1. Done!


the x-intercept(s) is/are where y=0, so we get {{{x^2+1 = 0}}}. Re-arranging, we get {{{x^2 = -1}}}, so x= +-{{{sqrt(-1)}}}. For this, you need to know Complex numbers. As far as you are concerned, there are no answers ie the curve does not cross the x-axis.


differentiate, to give dy/dx = 2x.


The turning point (vertex) is where the gradient (dy/dx) is zero, so ask this of th eequation: 2x=0 --> x=0. And y at this point?. It is y=1 (we already found the point earlier).


So, the vertex is at (0,1). The axis of symmetry always goes through the turning point, so axis is the y-axis, whose equation is x=0.


Just, so you can see it, here is the graph:


{{{graph(100,100,-4,4,0,10,x^2+1)}}}



jon.