1 of 8 0-day streak

Read and construct a gradient function from a function graph

Question 1 of 8

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The following exact data describe the function on the displayed horizontal scale

f(x)=-\left(x+3\right)\left(x-1\right)^2

The x-intercepts are \(x=-3,\quad 1\)

The relevant stationary or corner x-values are \(x=-\frac{5}{3},\quad 1\), with points \((-\frac{5}{3},-\frac{256}{27}),\quad (1,0)\)

From left to right, the function has negative slope on x < -5/3; positive slope on -5/3 < x < 1; negative slope on 1 < x

(a)1 mark

Determine where the function rises and falls, then select the corresponding gradient signs

The gradient is zero when \(x=-\frac{5}{3},\quad 1\)

f'(x)>0\text{ for }x<-\frac{5}{3}\quad f'(x)<0\text{ for }-\frac{5}{3}<x<1\quad f'(x)>0\text{ for }1<x

The gradient is zero when \(x=-\frac{5}{3},\quad 1\)

f'(x)<0\text{ for }x<-\frac{5}{3}\quad f'(x)>0\text{ for }-\frac{5}{3}<x<1\quad f'(x)<0\text{ for }1<x

The gradient has degree 3

The gradient is zero when \(x=-3,\quad 1\)

f'(x)<0\text{ for }x<-3\quad f'(x)>0\text{ for }-3<x<1\quad f'(x)<0\text{ for }1<x

The gradient is zero when \(x=-\frac{5}{3},\quad 1\)

f'(x)<0\text{ for }x<-\frac{5}{3}\quad f'(x)>0\text{ for }-\frac{5}{3}<x<1\quad f'(x)<0\text{ for }1<x

(b)1 mark

Enter the x-coordinate of the first listed turning or corner point

x-coordinate of first feature point