Graph of a function

It tells you that when the input is x, the output is y — i.e. f(x) = y. The x-axis shows inputs; the y-axis shows outputs.

f(3) = 7. Read the y-value at x = 3 directly from the graph.

Locate x = 4 on the horizontal axis, go straight up to the curve, then read across to the y-axis. That y-value is f(4).

f(0) = −5. The point (0, −5) is on the graph, so when x = 0 the output is −5.

Shape of the curve, any x- and y-intercepts, turning points (if present), and asymptotes (if relevant). Label key values. Accuracy matters less than the correct shape and labelled features.

Straight line → linear. U-shape → quadratic. J-curve → exponential. Wave → sinusoidal.

y-intercept at (0, 6). Gradient = −2: from (0, 6), go right 1 and down 2 to reach (1, 4). Draw a straight line through both points and label the y-intercept.

a < 0. The parabola has a maximum (peak) at the vertex. If a > 0 it opens upward with a minimum.

Go to x = 2 on the horizontal axis, read straight up to the curve, then across to the y-axis. Write the y-value you find. "Write down" means no working is needed.

Draw a horizontal line at y = 5. Where it meets the curve, read straight down to the x-axis. There may be more than one solution.

The function has two x-intercepts (zeros/roots) at x = −1 and x = 3. The curve crosses the x-axis at those points.

Printed graphs have limited precision. As long as your reading is within 0.2 of the true value, the mark is awarded. Always read as carefully as possible.

Exponential: approaches a horizontal asymptote (y → 0 as x → −∞), never crosses the x-axis (if a > 0). Quadratic: has a vertex (turning point), usually has two x-intercepts, is symmetric.

A horizontal asymptote at y = 4. The curve gets arbitrarily close but never equals 4.

A cubic polynomial or a sinusoidal function. A quadratic has only one turning point; two suggests a higher-degree polynomial or a periodic function.

y → ∞. The graph grows without bound — steeper and steeper. As x → −∞, y → 0 (horizontal asymptote).

x-intercept: where the graph crosses the x-axis — this is where y = 0. y-intercept: where the graph crosses the y-axis — this is where x = 0.

No. A function produces exactly one output for x = 0, so there is exactly one y-intercept. However, a function can have zero, one, or many x-intercepts.

The curve stays entirely above or below the x-axis — its output is never zero.

All three refer to the values of x where f(x) = 0 — i.e. where the graph meets the x-axis. They are the same thing.

Substitute x = 0 into the function and calculate the output. The y-intercept is at the point (0, f(0)).

f(0) = 0 − 0 + 7 = 7. y-intercept is (0, 7).

f(0) = 5 · 2⁰ = 5 · 1 = 5. y-intercept is (0, 5). For any exponential y = a · bˣ, the y-intercept is always (0, a).

When x = 0: y = m(0) + c = c. So the line always meets the y-axis at the constant term.

Set f(x) = 0 and solve. Each solution is an x-intercept (root/zero).

Set x² − x − 6 = 0. Factor: (x − 3)(x + 2) = 0. So x = 3 or x = −2. x-intercepts are (3, 0) and (−2, 0).

The x-values where f(x) = 0, typically as coordinates: e.g. (−2, 0) and (3, 0), or just x = −2 and x = 3. Using the GDC Zero function is fine.

No real x-intercepts — the parabola is entirely above or below the x-axis. The equation has no real solutions.

Times when h = 0 — i.e. when the ball is on the ground: t = 0 (launch) and t = 4 (lands). x-intercepts are times, not heights.

P(0) = 800. The y-intercept is the initial population of 800 (at time t = 0).

State what the y-intercept value represents using the context's real-world units and language. E.g. "800 is the initial population at the start of the study."

The fixed cost of 400 — even if n = 0 units are produced, the cost is still 400 (overhead/startup cost).

The range of x and y values displayed on screen. Set using Xmin, Xmax, Ymin, Ymax. If the window is wrong, key features of the graph will be off-screen.

The turning points are outside the default window. Zoom out — increase the x and y range (e.g. −15 to 15). Use ZoomFit or adjust Ymin/Ymax manually.

Key features (intercepts, turning points, asymptotes) may be off-screen in the default window. Missing them leads to incomplete or wrong answers.

Automatically adjusts the y-window to show all points of the graph within the current x-range. Use it when the default window shows nothing useful.

Graph the function. Use 2nd → Calc → Zero (TI-84). Set a left bound and right bound on either side of each zero. The GDC gives the exact x-value.

Graph both functions. Use 2nd → Calc → Intersect (TI-84). Move the cursor near the intersection and press Enter three times. The GDC gives both x and y coordinates.

The y-coordinate. Substitute x = 2.31 into either equation, or read y from the GDC screen. IB expects both coordinates: e.g. (2.31, 5.62).

Graph h(x) = f(x) − g(x) and find its zeros using the Zero function. Where h(x) = 0 is exactly where f(x) = g(x).

Graph f(x). Use 2nd → Calc → Maximum. Set a left bound before the peak and a right bound after it. The GDC returns both x and y coordinates of the maximum.

Both the x and y coordinates as a pair: e.g. (2, −3). Never write only the x-value — that loses the second mark.

Use Maximum for the peak and Minimum for the trough — run them separately with appropriate bounds around each turning point.

12. The maximum value is the y-coordinate of the turning point, not the x-coordinate.

Write both coordinates clearly: x = 3.46, y = 8.92 (3 s.f. unless told otherwise). Or write the coordinate pair (3.46, 8.92).

No — you must state the GDC result clearly (coordinates, equation, etc.) but no algebraic working is needed. Always write what you found, not how the GDC found it.

Paper 2 only. Paper 1 is the non-calculator paper. No GDC allowed on Paper 1.

3 significant figures (3 s.f.), unless the question specifies otherwise. Using more decimal places is not wrong but messy; using fewer can cost marks.