Question 41
$y=x^4-4x^3+4x^2-10$
$\frac{dy}{dx}=4x^3-12x^2+8x=0$
$x(x^2-3x+2)=0$
$x(x-1)(x-2)=0$
$x=0, 1$ or $2$
Co-ordinates of turning points are:
$(0,-10), (1,-9)$ and $(2,-10)$
Since the graph is a positive quartic, it must only cut the $x$-axis twice.
$y=x^4-4x^3+4x^2-10$
$\frac{dy}{dx}=4x^3-12x^2+8x=0$
$x(x^2-3x+2)=0$
$x(x-1)(x-2)=0$
$x=0, 1$ or $2$
Co-ordinates of turning points are:
$(0,-10), (1,-9)$ and $(2,-10)$
Since the graph is a positive quartic, it must only cut the $x$-axis twice.