Analysing Graphs with Differentiation

Because differentiation gives you the gradient of a curve, you can differentiate an equation to find turning points on a graph.

A turning point has a gradient of zero, so by differentiating the equation and setting $x$ to zero will give the $x$ -values of any turning points. Substituting $X$ into the original equation will give the $y$ -value for the coordinate.

By examining the gradient on either side of a turning point, you can determine whether the turning point was a maximum, minimum or a stationary value.

Example 1

For the equation $y = x^2+4x-2$ , find the turning point and indicate whether it is a maximum or a minimum.

$y$	$=x^2+4x-2$
$frac(text(d)y)(text(d)x)$	$=2x + 4$
Turning point is when	gradient equals zero
$0$	$=2x + 4$
$x$	$=-2$
Find y-coordinate
$y$	$=x^2+4x-2$
$y$	$=(-2)^2+4(-2)-2$
$y$	$=-6$
Coordinate is	(-2, -6)

Check the gradient on either side of the turning point:

Set $x$ = -3, gradient = 2(-3) + 4 = -2

Set $x$ = -1, gradient = 2(-1) + 4 = +2

Gradient is negative (going down) then zero then positive (going up): ∖ _ ∕

This is a minimum.

Answer: (-2, -6). It is a minimum.

Example 2

A cubic equation is given by $y = x^3 + 4x^2 - 3x$ . What is the coordinate of the turning point where $x < 0$ ?

$y$	$= x^3 + 4x^2 - 3x$
$frac(text(d)y)(text(d)x)$	$= 3x^2 + 8x - 3$
	Find the turning point where $frac(dy)(dx)=0$
0	$= 3x^2 + 8x - 3$
	$text(Use )x= frac(-b ±sqrt(b^2 - 4ac))(2a)$
	$= frac(-(8) ±sqrt((8)^2 - 4(3)(-3)))(2(3))$
	$= -3 and frac(1)(3)$
	Only interested in $x$ = -3
$y$	$=(-3)^3 + 4(-3)^2 - 3(-3)$
	$=18$

Answer: (-3, 18)