Question

Find the stationary point on the curve y=2x^2+x-1

y'=4x+1
For stationary points y'=0
4x+1=0
4x=-1
x=-0.25

When x=-0.25, y=4(-0.25)+1
y=0

Point = (-0.25, 0)

Can someone please tell me how the book gets (-0.25, -1.25)?

Answer 1

The point stationary point \((x,y)\) is of the original function. You plugged in the \(x\)-value into \(y'\) instead of \(y\). :)

Answer 2

You get \(x = -0.25\), so you plug it into \(y = 2x^2 + x - 1\).

Answer 3

That would return \(y = -1.125\).

So the stationary point is \((-0.25,-1.125)\)

Answer 4

Thanks, such a stupid error :P

Answer 5

I've made this error too, so you're certainly not alone!

Answer 6

*phew*