Curve Sketching 1
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Question 1 of 5
1. Question
Sketch the First Derivative Curve of the curve shown below:Hint
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A point that satisfies `f'(x)=0` is called a Stationary PointA positive `f'(x)` means a positive slopeA negative `f'(x)` means a negative slopeFirst, set up a separate plane with the `y` axis replaced with `f'(x)` axisNext, mark out the stationary points from the given curve into the revised graphNote that in the revised graph, stationary points are always on the `x` axisFor an easier transfer, mark out the movement of the given curveRemember that a positive or increasing slope means `f'(x)` is positive, and a negative or decreasing slope means `f'(x)` is negativeFinally, draw a curve on the revised graph depending on the signs marked on the original curve 
Question 2 of 5
2. Question
Sketch the curve:`f(x)=x^22x`Hint
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A point that satisfies `f'(x)=0` is called a Stationary PointA positive `f'(x)` means a positive slopeA negative `f'(x)` means a negative slopeFirst, equate `f'(x)` to `0` to find the stationary points`f(x)` `=` `x^22x` `f'(x)` `=` `2x2` `2x2` `=` `0` Equate to `0` `2(x1)` `=` `0` Factor out `2` `x` `=` `1` This means that there is a stationary point at `x=1`Find the corresponding `y` value of `x=1``f(x)` `=` `x^22x` $$f(\color{#007DDC}{1})$$ `=` $$\color{#007DDC}{1}^2)2(\color{#007DDC}{1})$$ `=` `12` `=` `1` Stationary Point: $$(\color{#007DDC}{1},\color{#e65021}{1})$$Next, graph `f'(x)=2x2` to identify where the curve increases or decreasesStart by setting up a plane with the `y` axis replaced with `f'(x)` axis and plot `x=1`Substitute `x=0` to know where the derivative graph intersects the `f'(x)` axis`f'(x)` `=` `2x2` `=` `2(0)2` Substitute `x=0` `=` `02` `f'(x)` `=` `2` Form a line by connecting the two pointsFrom the diagram above, we can say the following about the curve of `f(x)`:The slope is negative or decreasing at `x≤1`The slope is positive or increasing at `x≥1`Next, check where the curve intersects the `x` axis by equating `f(x)` to `0``f(x)` `=` `0` `x^22x` `=` `0` Equate to `0` `x(x2)` `=` `0` `x` `=` `0,2` The curve intersects `x=0` and `x=2`Lastly, draw a curve or a parabola over the plotted points, following the increase and decrease that have been set with the derivative graph 
Question 3 of 5
3. Question
Sketch the curve`f(x)=x^48x^2+16`Hint
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A point that satisfies `f'(x)=0` is called a Stationary PointA sign diagram is a way of visualizing a curve’s graph by indicating the inflection points, stationary points, and the increases or decreases in the curve.First, equate `f'(x)` to `0` to find the stationary points`f(x)` `=` `x^48x^2+16` `f'(x)` `=` `4x^316x=0` Equate `f'(x)` to `0` `=` `4x(x^24)=0` Factor out `4x` `=` `4x(x2)(x+2)=0` Factor out the difference of `2` squares `x=0,x=2,x=2` Find the corresponding `y` values by substituting each `x` value to the function`f(x)` `=` `x^48x^2+16` $$f(\color{#007DDC}{0})$$ `=` $$\color{#007DDC}{0}^48(\color{#007DDC}{0}^2)+16$$ `=` `00+16` `=` `16` Stationary Point: $$(\color{#007DDC}{0},\color{#e65021}{16})$$`f(x)` `=` `x^48x^2+16` $$f(\color{#007DDC}{2})$$ `=` $$\color{#007DDC}{2}^48(\color{#007DDC}{2}^2)+16$$ `=` `1632+16` `=` `0` Stationary Point:
$$(\color{#007DDC}{2},\color{#e65021}{0})$$`f(x)` `=` `x^48x^2+16` $$f(\color{#007DDC}{2})$$ `=` $$\color{#007DDC}{2}^48(\color{#007DDC}{2}^2)+16$$ `=` `1632+16` `=` `0` Stationary Point:
$$(\color{#007DDC}{2},\color{#e65021}{0})$$Next, create a sign diagram to identify where the curve increases or decreasesStart by setting up a horizontal line with matching indicators of the stationary pointsTest the gradient of a point to the left of `2` such as `x=3``f'(x)` `=` `4x^316x` $$f'(\color{#00880A}{3})$$ `=` $$4(\color{#00880A}{3})^316(\color{#00880A}{3})$$ `=` `4(27)+48` `=` `108+48` `=` `60` This value is negative, which means the curve’s slope at `3` is decreasingIndicate this on the sign diagram by adding a negative sign to the left of `2`Test the gradient of a point to the right of `2` such as `x=1``f'(x)` `=` `4x^316x` $$f'(\color{#00880A}{1})$$ `=` $$4(\color{#00880A}{1})^316(\color{#00880A}{1})$$ `=` `4(1)+16` `=` `4+16` `=` `12` This value is positive, which means the curve’s slope at `1` is increasingIndicate this on the sign diagram by adding a positive sign to the right of `2`Test the gradient of a point to the left of `2` such as `x=1``f'(x)` `=` `4x^316x` $$f'(\color{#00880A}{1})$$ `=` $$4(\color{#00880A}{1}^3)16(\color{#00880A}{1})$$ `=` `4(1)16` `=` `416` `=` `12` This value is negative, which means the curve’s slope at `1` is decreasingIndicate this on the sign diagram by adding a negative sign to the left of `2`Test the gradient of a point to the right of `2` such as `x=3``f'(x)` `=` `4x^316x` $$f'(\color{#00880A}{3})$$ `=` $$4(\color{#00880A}{3}^3)16(\color{#00880A}{3})$$ `=` `4(27)48` `=` `10848` `=` `60` This value is positive, which means the curve’s slope at `3` is increasingIndicate this on the sign diagram by adding a positive sign to the left of `2`Finally, draw a curve along the stationary points with the help of the sign diagram 
Question 4 of 5
4. Question
To help identify where the curve increases or decreases, sketch the first derivative of the curve:`f(x)=x^36x^2+9x+1`Hint
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A point that satisfies `f'(x)=0` is called a Stationary PointA positive `f'(x)` means a positive slopeA negative `f'(x)` means a negative slopeFirst, equate `f'(x)` to `0` to find the stationary points`f(x)` `=` `x^36x^2+9x+1` `f'(x)` `=` `3x^212x+9` `3x^212x+9` `=` `0` Equate to `0` `3(x^24x+3)` `=` `0` Factor out `3` `3(x3)(x1)` `=` `0` Factor out further `x` `=` `3,1` This means that there are stationary points at `x=3` and `x=1`To start graphing the derivative, set up a plane with the `y` axis replaced with `f'(x)` axisPlot `x=1` and `x=3` since this is where the derivative would intersect the `x` axisRecall that `f'(x)=3x^212x+9`Since the highest power is `2`, the graph would be a parabola. Draw this over the two plotted pointsFrom the diagram above, we can say the following about the curve of `f(x)`:The slope is positive at `x≤1`The slope is negative at `1≤x≤3`The slope is positive at `x≥3` 
Question 5 of 5
5. Question
Sketch the curve:`f(x)=x^36x^2+9x+1`Hint
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A point that satisfies `f'(x)=0` is called a Stationary PointA positive `f'(x)` means a positive slopeA negative `f'(x)` means a negative slopeFirst, equate `f'(x)` to `0` to find the stationary points`f(x)` `=` `x^36x^2+9x+1` `f'(x)` `=` `3x^212x+9` `3x^212x+9` `=` `0` Equate to `0` `3(x^24x+3)` `=` `0` Factor out `3` `3(x3)(x1)` `=` `0` Factor out further `x` `=` `3,1` This means that there are stationary points at `x=3` and `x=1`Find the corresponding `y` values by substituting each `x` value to the function`f(x)` `=` `x^36x^2+9x+1` $$f(\color{#007DDC}{3})$$ `=` $$\color{#007DDC}{3}^36(\color{#007DDC}{3}^2)+9(\color{#007DDC}{3})+1$$ `=` `2754+27+1` `=` `1` Stationary Point: $$(\color{#007DDC}{3},\color{#e65021}{1})$$`f(x)` `=` `x^36x^2+9x+1` $$f(\color{#007DDC}{1})$$ `=` $$\color{#007DDC}{1}^36(\color{#007DDC}{1}^2)+9(\color{#007DDC}{1})+1$$ `=` `16+9+1` `=` `5` Stationary Point: $$(\color{#007DDC}{1},\color{#e65021}{5})$$Next, graph `f'(x)=3x^212x+9` to identify where the curve increases or decreasesStart by setting up a plane with the `y` axis replaced with `f'(x)` axis and plot `x=1` and `x=3`Since the highest power of `f'(x)` is `2`, draw a parabola over the plotted pointsFrom the diagram above, we can say the following about the curve of `f(x)`:The slope is positive or increasing at `x≤1`The slope is negative or decreasing at `1≤x≤3`The slope is positive or increasing at `x≥3`Next, check where the curve intersects the `y` axis by substituting `x=0` to `f(x)``f(x)` `=` `x^36x^2+9x+1` `f(0)` `=` `0^36(0^2)+9(0)+1` Substitute `x=0` `=` `00+0+1` `y` `=` `1` The curve intersects `y=1`Lastly, draw a curve over the plotted points, following the increase and decrease that have been set with the derivative graph