INTRODUCTION
Graph lab
Everything covered in this lesson, in one place - useful for revision or printing.
\( \begin{gathered} y = a(x+p)^2 + q \\[4pt] y = \dfrac{a}{x+p} + q \\[4pt] y = a \cdot b^{x+p} + q \end{gathered} \)
Paper 1 has three Grade 11 graphs: the parabola, the hyperbola and the exponential. All three are a base graph changed by the same numbers: a stretches or flips it, p moves it left or right, and q moves it up or down.
The good news: p and q do exactly the same job in all three. Learn them once.
Pick a graph in the lab below and drag the sliders. The faint dashed curve is the base graph, so you can see what changed. The readout under the graph lists the turning point or asymptotes, intercepts, domain and range as you drag.
\( y = x^2 \)
Start every parabola from \(y = x^2\). Learn these points:
The graph is symmetrical about the y-axis (the line \(x = 0\)), and its lowest point, the turning point, is \((0;\,0)\). Every y-value is \(\ge 0\), so the range is \(y \ge 0\).
\( y = -2x^2 \)
a multiplies every y-value. Predict first, then drag a to check.
\( y = (x - 3)^2 \)
p sits inside the bracket with x. Ask one question: what x makes the bracket zero?
\( y = x^2 - 4 \)
q is added on at the end, so it moves every point up or down.
\( y = -(x - 1)^2 + 4 \)
In the form \(y = a(x+p)^2 + q\) you can read the turning point straight off: it is \((-p;\,q)\).
\( \begin{gathered} \text{TP } (2;\,-1) \\[2pt] \text{through } (0;\,3) \end{gathered} \)
A favourite exam question: you are given the turning point and one other point. Substitute the turning point into \(y = a(x+p)^2 + q\) first:
\( \begin{gathered} y = a(x+p)^2 + q \\[4pt] \text{TP } (-p;\,q) \\[2pt] \text{axis } x = -p \end{gathered} \)
• a > 0: opens upwards, minimum value q, range \(y \ge q\).
• a < 0: opens downwards, maximum value q, range \(y \le q\).
• y-intercept: put x = 0. x-intercepts: put y = 0 and solve.
• Domain: \(x \in \mathbb{R}\) — every parabola.
• Given \(y = ax^2 + bx + c\)? The axis of symmetry is \(x = -\dfrac{b}{2a}\), or complete the square to get the turning-point form.
\( y = \dfrac{1}{x} \)
You can never divide by 0, so \(x = 0\) is not allowed: the graph has a vertical asymptote at \(x = 0\). And \(\tfrac{1}{x}\) is never 0, so there is a horizontal asymptote at \(y = 0\). The graph gets closer and closer to these dashed lines but never touches them.
The two branches lie in the 1st and 3rd quadrants. The green dotted lines \(y = x\) and \(y = -x\) are its axes of symmetry.
\( y = \dfrac{1}{x} + 2 \)
q does exactly what it did to the parabola: it moves every point up or down. Watch the dashed lines as you drag.
\( y = \dfrac{1}{x - 2} \)
Same question as for the parabola: what x makes the bracket zero? For a hyperbola that x-value is forbidden.
\( y = \dfrac{-2}{x} \)
The sign of a decides which quadrants (measured from the asymptotes) the branches lie in.
\( y = \dfrac{2}{x - 1} + 3 \)
Both axes of symmetry pass through the point where the asymptotes cross, \((-p;\,q)\), with gradients 1 and −1:
\( y = \dfrac{2}{x - 1} + 3 \)
Intercepts work the same way for every graph.
• y-intercept: put \(x = 0\).
• x-intercept: put \(y = 0\): \(\dfrac{2}{x-1} = -3\), so \(x - 1 = -\tfrac{2}{3}\) and \(x = \tfrac{1}{3}\).
\( \begin{gathered} y = \dfrac{a}{x+p} + q \\[4pt] \text{asymptotes } x = -p, \ y = q \end{gathered} \)
• Domain: \(x \in \mathbb{R},\ x \ne -p\). Range: \(y \in \mathbb{R},\ y \ne q\).
• a > 0: 1st and 3rd quadrants; a < 0: 2nd and 4th.
• Axes of symmetry: \(y = x + p + q\) and \(y = -x - p + q\).
• Sketch the asymptotes first (dashed), then the intercepts, then the branches.
\( y = 2^x \)
The base b is a positive number other than 1. For \(y = 2^x\):
• Every exponential \(y = b^x\) passes through (0; 1), because \(b^0 = 1\).
• \(2^x\) is always positive, so the x-axis (\(y = 0\)) is a horizontal asymptote and the range is \(y > 0\).
\( y = \left(\tfrac{1}{2}\right)^x \)
Now change the base. Drag b after you predict.
\( y = 2^x - 3 \)
q moves the graph up or down, and the horizontal asymptote goes with it.
\( y = 2^{x+1} \)
p is in the exponent, with x. Bracket-zero again: \(x + 1 = 0\) gives \(x = -1\), so the graph moves 1 to the left.
\( y = -2^x + 4 \)
\(-2^x\) means \(-(2^x)\): a = −1 reflects the graph in its asymptote, so it now lies below it.
\( \begin{gathered} y = a \cdot b^{x+p} + q \\[4pt] \text{asymptote } y = q \end{gathered} \)
• b > 0, b ≠ 1. b > 1: increasing; 0 < b < 1: decreasing (if a > 0).
• a > 0: above the asymptote, range \(y > q\). a < 0: below it, range \(y < q\).
• Domain: \(x \in \mathbb{R}\). There is no vertical asymptote.
• Range uses > or <, never ≥ or ≤: the asymptote value is never reached.
\( \begin{gathered} \text{Average gradient} \\[2pt] = \dfrac{f(x_2) - f(x_1)}{x_2 - x_1} \end{gathered} \)
The average gradient between two points on a curve is the gradient of the straight line (the secant) joining them. Here \(f(x) = x^2 - 2x - 3\).
Drag the points A and B along the curve, or use the sliders.
\( f(x) = x^2 - 2x - 3 \)
The secant can be horizontal. Think about where the axis of symmetry of f is before you drag.
\( m_{\text{avg}} = \dfrac{y_2 - y_1}{x_2 - x_1} \)
• Find both y-values from the equation first, then use the gradient formula.
• Positive: the curve is higher at B. Negative: lower at B. Zero: the same height.
• It works for every graph — parabola, hyperbola or exponential.
• Order does not matter, as long as you subtract in the same order on top and bottom.
Move A and B freely: watch how the gradient changes as the secant swings.
\( \begin{gathered} f(x) = -x^2 + 2x + 3 \\[2pt] g(x) = x + 1 \end{gathered} \)
Every “for which values of x” question is built from the x-values where the graphs meet, or cut the x-axis. Always find them first.
To find where they meet, make the equations equal: \(f(x) = g(x)\).
\( f(x) > g(x) \)
\(f(x) > g(x)\) asks: where is the graph of f above the graph of g? Read it between the intersections.
\( f(x) \cdot g(x) \le 0 \)
A product is negative when one factor is positive and the other negative. On the graph: one graph above the x-axis and the other below it. Because of the ≤, the product may also be 0, so the x-intercepts of either graph are included.
f cuts the x-axis at \(x = -1\) and \(x = 3\); g cuts it at \(x = -1\).
\( x \cdot f(x) \ge 0 \)
Here the second factor is just x. x is positive to the right of the y-axis and negative to the left. So you need:
• x ≥ 0 and f(x) ≥ 0 (right of the y-axis, f on or above the x-axis), or
• x ≤ 0 and f(x) ≤ 0 (left of the y-axis, f on or below the x-axis).
\( \begin{gathered} \text{length} = f(x) - g(x) \\[2pt] = -x^2 + x + 2 \end{gathered} \)
For \(-1 \le x \le 2\), f is above g, so the length of a vertical line between the graphs is top minus bottom: \(f(x) - g(x)\). That is a new parabola, so its maximum is at its turning point.
Drag the green line (or use the slider) to find the longest one.
\( \begin{gathered} \;\to\; \text{ends excluded} \\[4pt] \le,\ \ge \;\to\; \text{ends included} \end{gathered} \)
\[ \begin{gathered} p\text{: } x + p = 0 \Rightarrow x = -p \\[4pt] q\text{: up or down} \\[4pt] a < 0\text{: flips} \end{gathered} \]
The mistakes to avoid: