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Functions - Linear, quadratic, exponential, and logarithmic functions

Grade 10IB

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

🔑Concepts

Linear functions represent a constant rate of change and form a straight-line graph. Visually, the slope mm determines the steepness—a positive slope rises from left to right, while a negative slope falls. The yy-intercept cc is the point (0,c)(0, c) where the line crosses the vertical axis, and the 'rise over run' staircase pattern describes how to move between points on the line.

Quadratic functions are defined by a squared variable and form a U-shaped curve called a parabola. The graph is perfectly symmetrical about a vertical line called the axis of symmetry. If the leading coefficient a>0a > 0, the parabola opens upwards like a cup (minimum point); if a<0a < 0, it opens downwards like an umbrella (maximum point).

The vertex of a quadratic function is the highest or lowest point on the graph. In vertex form y=a(xh)2+ky = a(x-h)^2 + k, the point (h,k)(h, k) explicitly identifies this turning point. Visually, changing hh slides the parabola horizontally, while changing kk shifts it vertically.

Exponential functions involve a constant base raised to a variable power, such as y=abxy = ab^x. These functions model rapid growth (when b>1b > 1) or decay (when 0<b<10 < b < 1). The graph features a horizontal asymptote, usually the xx-axis (y=0y=0), which the curve approaches but never touches, creating a characteristic L-shaped curve.

Logarithmic functions are the inverse of exponential functions, written as y=logb(x)y = \log_b(x). Graphically, a log function is a reflection of an exponential function across the diagonal line y=xy = x. It features a vertical asymptote at x=0x = 0 and passes through the point (1,0)(1, 0), growing very slowly as xx increases.

The domain of a function is the set of all possible xx-values (inputs), while the range is the set of all possible yy-values (outputs). For linear functions, both are typically all real numbers. For quadratics, the range is restricted by the vertex yy-coordinate. For the parent exponential function y=bxy = b^x, the range is y>0y > 0, while for the logarithmic function y=logb(x)y = \log_b(x), the domain is x>0x > 0.

Intercepts are where the graph crosses the axes. To find the yy-intercept, set x=0x = 0 and solve for yy. To find the xx-intercepts (also known as roots, zeros, or solutions), set y=0y = 0 and solve for xx. For a quadratic, there can be zero, one, or two xx-intercepts depending on the discriminant.

📐Formulae

Linear (Slope-Intercept Form): y=mx+cy = mx + c

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

Quadratic (Standard Form): y=ax2+bx+cy = ax^2 + bx + c

Quadratic (Vertex Form): y=a(xh)2+ky = a(x - h)^2 + k

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

Axis of Symmetry: x=b2ax = -\frac{b}{2a}

Exponential Form: y=abx+ky = a \cdot b^x + k

Logarithmic Identity: y=logb(x)    by=xy = \log_b(x) \iff b^y = x

💡Examples

Problem 1:

Given the quadratic function f(x)=x26x+5f(x) = x^2 - 6x + 5, find the coordinates of the vertex and the xx-intercepts.

Solution:

  1. Find the xx-coordinate of the vertex using x=b2ax = -\frac{b}{2a}: x=62(1)=3x = -\frac{-6}{2(1)} = 3. \n2. Substitute x=3x = 3 into the function to find the yy-coordinate: f(3)=(3)26(3)+5=918+5=4f(3) = (3)^2 - 6(3) + 5 = 9 - 18 + 5 = -4. So, the vertex is (3,4)(3, -4). \n3. To find xx-intercepts, set f(x)=0f(x) = 0: x26x+5=0x^2 - 6x + 5 = 0. \n4. Factor the quadratic: (x5)(x1)=0(x - 5)(x - 1) = 0. \n5. Solve for xx: x=5x = 5 and x=1x = 1. The intercepts are (1,0)(1, 0) and (5,0)(5, 0).

Explanation:

We used the axis of symmetry formula to locate the center of the parabola (the vertex) and used factoring to find the points where the curve crosses the horizontal axis.

Problem 2:

An exponential function is given by f(x)=32x+kf(x) = 3 \cdot 2^x + k. If the graph passes through the point (2,15)(2, 15), determine the value of kk and the equation of the horizontal asymptote.

Solution:

  1. Substitute the point (2,15)(2, 15) into the equation: 15=322+k15 = 3 \cdot 2^2 + k. \n2. Simplify the exponent: 15=34+k15 = 3 \cdot 4 + k. \n3. Solve for kk: 15=12+k    k=315 = 12 + k \implies k = 3. \n4. The function is f(x)=32x+3f(x) = 3 \cdot 2^x + 3. \n5. For an exponential function y=abx+ky = ab^x + k, the horizontal asymptote is always y=ky = k. Therefore, the asymptote is y=3y = 3.

Explanation:

By plugging in a known point, we solved for the vertical translation constant kk. This kk value also determines the level at which the graph flattens out (the asymptote).