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Functions - Linear, quadratic, exponential, and logarithmic functions

Grade 10IB

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

πŸ”‘Concepts

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Linear functions represent a constant rate of change and form a straight-line graph. Visually, the slope mm determines the steepnessβ€”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.

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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).

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The vertex of a quadratic function is the highest or lowest point on the graph. In vertex form y=a(xβˆ’h)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.

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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.

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Logarithmic functions are the inverse of exponential functions, written as y=log⁑b(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.

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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=log⁑b(x)y = \log_b(x), the domain is x>0x > 0.

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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.

πŸ“Formulae

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

Slope Formula: m=y2βˆ’y1x2βˆ’x1m = \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(xβˆ’h)2+ky = a(x - h)^2 + k

Quadratic Formula: x=βˆ’bΒ±b2βˆ’4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

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

Exponential Form: y=aβ‹…bx+ky = a \cdot b^x + k

Logarithmic Identity: y=log⁑b(x)β€…β€ŠβŸΊβ€…β€Šby=xy = \log_b(x) \iff b^y = x

πŸ’‘Examples

Problem 1:

Given the quadratic function f(x)=x2βˆ’6x+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=βˆ’b2ax = -\frac{b}{2a}: x=βˆ’βˆ’62(1)=3x = -\frac{-6}{2(1)} = 3. \n2. Substitute x=3x = 3 into the function to find the yy-coordinate: f(3)=(3)2βˆ’6(3)+5=9βˆ’18+5=βˆ’4f(3) = (3)^2 - 6(3) + 5 = 9 - 18 + 5 = -4. So, the vertex is (3,βˆ’4)(3, -4). \n3. To find xx-intercepts, set f(x)=0f(x) = 0: x2βˆ’6x+5=0x^2 - 6x + 5 = 0. \n4. Factor the quadratic: (xβˆ’5)(xβˆ’1)=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)=3β‹…2x+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=3β‹…22+k15 = 3 \cdot 2^2 + k. \n2. Simplify the exponent: 15=3β‹…4+k15 = 3 \cdot 4 + k. \n3. Solve for kk: 15=12+kβ€…β€ŠβŸΉβ€…β€Šk=315 = 12 + k \implies k = 3. \n4. The function is f(x)=3β‹…2x+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).