Analyse sideways-parabola and root-function families

What should be decided first for the relation below?

Change h in the real cube-root family y=∛(x−h)+2. The table uses parent inputs that are perfect cubes, including negative ones

What happens as h increases by one?

Reading features of power relations

Power relations can describe a full curve or a selected function branch. Find the defining point and orientation, then apply the correct radicand condition and branch sign to determine domain and outputs

1. 1Identify the defining point The constants that shift the relation identify its vertex or central point before the orientation is read \(\displaystyle \left(y-k\right)^2=a\left(x-h\right)\quad\Rightarrow\quad (h,k)\)
2. 2Check a square-root domain For a square-root branch, the radicand must be non-negative, including zero \(\displaystyle y=-1+2\sqrt{x-2}\quad\Rightarrow\quad x-2\ge0\quad\Rightarrow\quad x\in[2,\infty)\)
3. 3Distinguish a full relation from a branch A full sideways-parabola relation has two branches, while one chosen sign gives one branch and can define a function \(\displaystyle \left(y+1\right)^2=4\left(x-2\right)\quad\Rightarrow\quad y=-1\pm2\sqrt{x-2}\)
4. 4Use the correct real domain Cube-root expressions accept every real input, unlike square-root expressions that require a non-negative radicand \(\displaystyle y=\sqrt[3]{x-2}\quad\Rightarrow\quad D=\mathbb{R}\)

Relation and input

\(\displaystyle (y-1)^2=2(x+3),\quad x=-1\)

For the displayed relation and x=−1, calculate the radicand in its square-root form

Relation and input

\(\displaystyle (y-1)^2=2(x+3),\quad x=-1\)

Using the upper branch, calculate y when x=−1.

Treating the full relation as one branch. One of these lines is right. Which one?

Build a full drawing from the existing features

Use exact parent points, algebra, domain restrictions and symmetry; differentiation is not required. Label every allowed branch, intercept, endpoint and asymptote, then compare the complete drawing with the feedback

1. 1Prepare the whole drawing State the domain and range. Label axes and scale; find intercepts, endpoints, asymptotes and exact key points before joining anything
2. 2Show the correct connections Draw every allowed branch with the correct direction and shape. Use open/closed endpoints, leave gaps at excluded inputs, and show end behaviour. A handful of correct dots is not a complete sketch
3. 3Check the mathematics without calculus Substitute labelled points, use factors and signs between roots, compare symmetry and both ends, and use tables or simple inequalities to check the shape. Distinguish a local extremum from a bound on the whole domain. A plotted model is feedback, not your submitted drawing

An endpoint is not an asymptote

The radicand requires x ≥ −1. Begin with the filled endpoint (−1,3). Exact points are (0,1), (3,−1) and (8,−3). The curve decreases, becoming less steep, and has range y ≤ 3. Its x-intercept is (5/4,0). It has no vertical or horizontal asymptote; do not extend it left of its endpoint

1. 1Find the x-intercept Set the output to zero, isolate the square root, then square both sides \(\displaystyle 0=3-2\sqrt{x+1}\quad\Rightarrow\quad\sqrt{x+1}=\frac{3}{2}\quad\Rightarrow\quad x+1=\frac{9}{4}\quad\Rightarrow\quad x=\frac{5}{4}\)

Rule or relation

\(\displaystyle f(x)=3-2\sqrt{x+1}\)

Find the exact x-coordinate of the x-intercept by solving the displayed equation

Your complete graph construction

Sketch g(x)=2√(x−3)−4 on its maximal domain. State domain and range, mark the included endpoint and every intercept, and label at least two further exact points. Describe which way the curve extends and explain why an endpoint is not an asymptote. Use parent square-root points rather than differentiation

Review the whole drawing and its domain

Domain [3,∞), range [−4,∞); filled endpoint (3,−4), x-intercept (7,0), and points (4,−2),(12,2),(19,4). There is no y-intercept because zero is outside the domain. The curve increases to the right and becomes less steep. There is no vertical or horizontal asymptote; do not draw a second lower branch

A real cube root has both negative and positive inputs

The real cube root allows every real input. Plot (−7,0),(0,1), the centre (1,2),(2,3) and (9,4). Join one continuous increasing curve through the centre, where it is very steep, not a pair of disconnected branches. Domain and range are R. It has centre symmetry about (1,2) and no vertical or horizontal asymptote. The graph changes its bending direction at the centre; no differentiation is needed to obtain the key points

Rule or relation

\(\displaystyle f(x)=\sqrt[3]{x-1}+2\)

Find f(−7) using the real cube root

Your complete graph construction

Sketch g(x)=2−∛(x+1) by hand using parent inputs that are perfect cubes. Label the centre, both intercepts and at least two further exact points, state domain and range, and show both end directions. Explain why negative inputs inside the cube root are allowed and why the steep centre is not a vertical asymptote

Review the whole drawing and its domain

Domain and range R. The centre is (−1,2), x-intercept (7,0), y-intercept (0,1), and further points (−9,4),(−2,3). The graph is decreasing, with centre symmetry about (−1,2); the left end rises without bound and the right end falls without bound. It passes continuously through the centre and has no asymptotes. The real cube root of a negative number is negative, unlike a real square root

Draw the full sideways relation, not one branch

The square-root forms are y=1±√(2(x+3)), so x≥−3. Start both branches at the vertex (−3,1). Plot the paired points (−1,3),(−1,−1),(5,5),(5,−3), and join a smooth sideways parabola opening right. The x-intercept is (−5/2,0); the y-intercepts are (0,1±√6). Domain [−3,∞), range R, with symmetry about y=1. At x=−1 there are two outputs, so the full relation is not a function of x. Its upper and lower branches separately are functions. There are no vertical or horizontal asymptotes

1. 1Find the x-intercept Set the output to zero \(\displaystyle y=0\quad\Rightarrow\quad 1=2(x+3)\quad\Rightarrow\quad x=-\frac{5}{2}\)
2. 2Find the y-intercepts Set the input to zero and take both square-root signs \(\displaystyle x=0\quad\Rightarrow\quad(y-1)^2=6\quad\Rightarrow\quad y=1\pm\sqrt{6}\)

Rule or relation

\(\displaystyle (y-1)^2=2(x+3)\)

How many distinct outputs does the full relation have at x=−1?

Your complete graph construction

Draw the complete relation (y+2)²=4(1−x), marking its vertex, all intercepts, paired exact points, symmetry and direction of opening. State its domain and range and use an explicit vertical-line example to decide whether it is a function of x. On separate axes draw only the branch y=−2+2√(1−x); compare its range with the full relation. Include the vertex in both drawings

Review the whole drawing and its domain

The full relation opens left from (1,−2), with domain (−∞,1], range R and symmetry y=−2. Its x-intercept is (0,0); its two y-intercepts are (0,0),(0,−4). Other paired points are (−3,2),(−3,−6). At x=0 the two distinct outputs 0 and −4 prove it is not a function of x. The upper branch has the same domain, range [−2,∞), and only the intercept (0,0). Neither the full relation nor its branch has a vertical or horizontal asymptote