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Algebra and Graphs - Linear, quadratic, and exponential graphs

Grade 11A Level

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

🔑Concepts

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Linear graphs represent a constant rate of change. The equation y=mx+cy = mx + c defines a straight line where mm is the gradient (slope) and cc is the yy-intercept. A positive mm slopes upwards from left to right, while a negative mm slopes downwards.

A linear graph showing a line with a positive gradient and y-intercept at (0,1).
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Quadratic graphs follow the form y=ax2+bx+cy = ax^2 + bx + c and create a 'U' or 'n' shaped curve called a parabola. If a>0a > 0, the curve opens upwards; if a<0a < 0, it opens downwards. Key features include the vertex (turning point) and the yy-intercept (at x=0x=0).

A quadratic graph showing an upward-opening parabola with a vertex at y=2.
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Exponential graphs have the form y=axy = a^x or y=k⋅axy = k \cdot a^x where a>0a > 0. These graphs grow or decay rapidly and never cross the x-axis (y=0y=0 is an asymptote). If a>1a > 1, the graph shows growth; if 0<a<10 < a < 1, it shows decay.

An exponential growth graph curving upwards from the x-axis.
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Perpendicular lines intersect at right angles. Their gradients m1m_1 and m2m_2 are negative reciprocals of each other, meaning m1⋅m2=−1m_1 \cdot m_2 = -1. For example, if a line has a gradient of 22, the perpendicular line has a gradient of −12-\frac{1}{2}.

📐Formulae

m=y2−y1x2−x1m = \frac{y_2 - y_1}{x_2 - x_1} (Gradient formula)

y=mx+cy = mx + c (Slope-intercept form)

x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} (Quadratic formula to find x-intercepts)

x=−b2ax = -\frac{b}{2a} (X-coordinate of the vertex of a quadratic)

m1⋅m2=−1m_1 \cdot m_2 = -1 (Condition for perpendicular lines)

💡Examples

Problem 1:

Find the equation of the line passing through the points (2,5)(2, 5) and (4,9)(4, 9).

Solution:

y=2x+1y = 2x + 1

Explanation:

First, calculate the gradient: m=(9−5)/(4−2)=4/2=2m = (9 - 5) / (4 - 2) = 4 / 2 = 2. Substitute m=2m=2 and point (2,5)(2, 5) into y=mx+cy = mx + c: 5=2(2)+c⇒5=4+c⇒c=15 = 2(2) + c \Rightarrow 5 = 4 + c \Rightarrow c = 1. Thus, y=2x+1y = 2x + 1.

Problem 2:

Find the coordinates of the turning point for the quadratic graph y=x2−6x+5y = x^2 - 6x + 5.

Solution:

(3,−4)(3, -4)

Explanation:

The x-coordinate of the vertex is x=−b/(2a)=−(−6)/(2⋅1)=3x = -b / (2a) = -(-6) / (2 \cdot 1) = 3. Substitute x=3x = 3 back into the original equation to find yy: y=(3)2−6(3)+5=9−18+5=−4y = (3)^2 - 6(3) + 5 = 9 - 18 + 5 = -4.

Problem 3:

Determine the value of kk for the exponential graph y=k⋅2xy = k \cdot 2^x if it passes through the point (3,40)(3, 40).

Solution:

k=5k = 5

Explanation:

Substitute the coordinates (x,y)=(3,40)(x, y) = (3, 40) into the equation: 40=k⋅2340 = k \cdot 2^3. Since 23=82^3 = 8, the equation becomes 40=8k40 = 8k. Dividing both sides by 8 gives k=5k = 5.

Problem 4:

Identify the xx-intercepts of the quadratic function y=x2−4y = x^2 - 4.

Graph of y = x^2 - 4 showing intersections with the x-axis at -2 and 2.

Solution:

  1. To find xx-intercepts, set y=0y = 0: 0=x2−40 = x^2 - 4.
  2. Factor the difference of squares: (x−2)(x+2)=0(x - 2)(x + 2) = 0.
  3. Solve for xx: x=2x = 2 or x=−2x = -2.

Explanation:

The xx-intercepts are the points where the curve crosses the horizontal axis. For a quadratic of the form y=x2−k2y = x^2 - k^2, the intercepts are always ±k\pm k.

Problem 5:

A curve has the equation y=2x−3y = 2^x - 3. (a) Find the yy-intercept of the graph. (b) Identify the equation of the horizontal asymptote. (c) Sketch the graph of the function showing its behavior as xx increases and decreases.

Graph of the exponential function y = 2^x - 3 showing the horizontal asymptote at y = -3 and y-intercept at -2.

Solution:

(a) To find the yy-intercept, set x=0x = 0: y=20−3y = 2^0 - 3 y=1−3y = 1 - 3 y=−2y = -2 The yy-intercept is (0,−2)(0, -2).

(b) As x→−∞x \to -\infty, 2x→02^x \to 0. Therefore, y→0−3y \to 0 - 3. The horizontal asymptote is y=−3y = -3.

(c) As xx increases, 2x2^x grows exponentially. When x=2x=2, y=22−3=1y = 2^2 - 3 = 1. When x=3x=3, y=23−3=5y = 2^3 - 3 = 5.

Explanation:

Exponential graphs of the form y=ax+ky = a^x + k have a horizontal asymptote at y=ky = k. The yy-intercept is found by evaluating the function at x=0x = 0. Since a>1a > 1, the graph shows exponential growth, rising steeply as xx becomes more positive.