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Functions - Function notation and graphing

Grade 10IB

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

Function Definition and Vertical Line Test: A function is a relation where every input xx has exactly one output yy. Visually, a graph represents a function if any vertical line drawn through it intersects the curve at most once. If a vertical line hits the graph twice, it is a relation but not a function.

Function Notation: We use f(x)f(x) to denote the output of a function for a given input xx. For example, if f(x)=2x+3f(x) = 2x + 3, then f(4)f(4) means substituting 44 for xx. On a graph, the xx-coordinate is the input, and the yy-coordinate (f(x)f(x)) represents the vertical height of the point.

Domain and Range: The domain is the set of all possible input values (xx-axis), while the range is the set of all possible output values (yy-axis). On a graph, the domain is the horizontal 'spread' from left to right, and the range is the vertical 'spread' from the lowest point to the highest point.

Linear Functions: These take the form f(x)=mx+cf(x) = mx + c. The graph is a straight line where mm is the gradient (slope) and cc is the yy-intercept. A positive mm results in a line sloping upwards to the right, while a negative mm slopes downwards.

Quadratic Functions: Written as f(x)=ax2+bx+cf(x) = ax^2 + bx + c, these form a symmetrical curve called a parabola. If a>0a > 0, the parabola opens upwards (U-shape) with a minimum point; if a<0a < 0, it opens downwards (n-shape) with a maximum point called the vertex.

Intercepts on a Graph: The yy-intercept occurs where the graph crosses the vertical axis, found by calculating f(0)f(0). The xx-intercepts (or zeros) are where the graph crosses the horizontal axis, found by solving the equation f(x)=0f(x) = 0.

Composite Functions: This involves placing one function inside another, denoted as (fg)(x)(f \circ g)(x) or f(g(x))f(g(x)). Visually, this means the output of the 'inner' function gg becomes the input for the 'outer' function ff.

📐Formulae

f(x)=yf(x) = y

m=y2y1x2x1m = \frac{y_2 - y_1}{x_2 - x_1}

f(x)=mx+cf(x) = mx + c

f(x)=a(xh)2+kf(x) = a(x - h)^2 + k

x=b±b24ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

(fg)(x)=f(g(x))(f \circ g)(x) = f(g(x))

💡Examples

Problem 1:

Given the function f(x)=3x212x+9f(x) = 3x^2 - 12x + 9, find the yy-intercept and the xx-intercepts of the graph.

Solution:

Step 1: To find the yy-intercept, evaluate f(0)f(0). f(0)=3(0)212(0)+9=9f(0) = 3(0)^2 - 12(0) + 9 = 9. The yy-intercept is at (0,9)(0, 9). \nStep 2: To find the xx-intercepts, set f(x)=0f(x) = 0. 3x212x+9=03x^2 - 12x + 9 = 0. \nStep 3: Factor out the common factor of 33: 3(x24x+3)=03(x^2 - 4x + 3) = 0. \nStep 4: Factor the quadratic: 3(x3)(x1)=03(x - 3)(x - 1) = 0. \nStep 5: Solve for xx: x=3x = 3 or x=1x = 1. The xx-intercepts are at (1,0)(1, 0) and (3,0)(3, 0).

Explanation:

The yy-intercept is found by setting the input to zero. The xx-intercepts are the roots of the equation where the output is zero.

Problem 2:

If f(x)=2x+5f(x) = 2x + 5 and g(x)=x2g(x) = x^2, find the value of f(g(3))f(g(3)).

Solution:

Step 1: Evaluate the inner function g(3)g(3). g(3)=32=9g(3) = 3^2 = 9. \nStep 2: Substitute this result into the outer function ff. f(9)=2(9)+5f(9) = 2(9) + 5. \nStep 3: Calculate the final value: 18+5=2318 + 5 = 23. So, f(g(3))=23f(g(3)) = 23.

Explanation:

This is a composite function problem. You must process the innermost operation first and use that result as the input for the next function.