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Algebra - Graphing linear and quadratic functions

Grade 9IB

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

πŸ”‘Concepts

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Linear Functions and Slope-Intercept Form: A linear function is represented by the equation y=mx+cy = mx + c, where the graph is a straight line. The value mm represents the gradient or slope; if m>0m > 0, the line rises from left to right, and if m<0m < 0, it falls. The constant cc is the y-intercept, the exact point (0,c)(0, c) where the line crosses the vertical y-axis.

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Finding Intercepts: To find where a graph crosses the axes, set the opposite variable to zero. For the x-intercept (where the graph touches the horizontal axis), set y=0y = 0 and solve for xx. For the y-intercept (where the graph touches the vertical axis), set x=0x = 0 and solve for yy. This applies to both linear lines and quadratic curves.

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Quadratic Functions and the Parabola: Quadratic functions follow the form y=ax2+bx+cy = ax^2 + bx + c and create a symmetrical 'U' shaped curve called a parabola. If aa is positive, the parabola opens upwards (like a smile) and has a minimum point; if aa is negative, it opens downwards (like a frown) and has a maximum point.

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The Vertex and Axis of Symmetry: The vertex is the 'turning point' of the parabola, representing its highest or lowest point. Every parabola is perfectly symmetrical across a vertical line called the Axis of Symmetry, which passes through the vertex. Visually, this line acts like a mirror, splitting the curve into two identical halves.

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Roots and Zeros: The solutions to a quadratic equation ax2+bx+c=0ax^2 + bx + c = 0 are the x-intercepts of the graph, also known as roots or zeros. A parabola can cross the x-axis at two points (two real roots), touch it at exactly one point (one repeated root), or not cross it at all (no real roots), appearing to 'float' above or below the axis.

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Gradient (Slope) Calculation: The gradient measures the 'steepness' of a line. It is visually understood as the 'rise over run'β€”the vertical change divided by the horizontal change between any two points (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) on the line.

πŸ“Formulae

y=mx+cy = mx + c

m=y2βˆ’y1x2βˆ’x1m = \frac{y_2 - y_1}{x_2 - x_1}

y=ax2+bx+cy = ax^2 + bx + c

x=βˆ’b2ax = -\frac{b}{2a}

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

πŸ’‘Examples

Problem 1:

Determine the gradient and y-intercept of the line 2yβˆ’4x=62y - 4x = 6, then find the x-intercept.

Solution:

  1. Rewrite in y=mx+cy = mx + c form: 2y=4x+6β€…β€ŠβŸΉβ€…β€Šy=2x+32y = 4x + 6 \implies y = 2x + 3.
  2. Identify mm and cc: Gradient m=2m = 2, y-intercept c=3c = 3.
  3. Find x-intercept by setting y=0y = 0: 0=2x+3β€…β€ŠβŸΉβ€…β€Šβˆ’3=2xβ€…β€ŠβŸΉβ€…β€Šx=βˆ’1.50 = 2x + 3 \implies -3 = 2x \implies x = -1.5.

Explanation:

To analyze a linear graph, it is easiest to convert the equation to slope-intercept form. The x-intercept is found by solving for the point where the line crosses the horizontal axis (y=0y=0).

Problem 2:

For the quadratic function y=x2βˆ’6x+5y = x^2 - 6x + 5, find the coordinates of the vertex and the x-intercepts.

Solution:

  1. Find Axis of Symmetry (xx coordinate of vertex): x=βˆ’b2a=βˆ’βˆ’62(1)=3x = -\frac{b}{2a} = -\frac{-6}{2(1)} = 3.
  2. Find yy coordinate of vertex: Substitute x=3x = 3 into the equation: y=(3)2βˆ’6(3)+5=9βˆ’18+5=βˆ’4y = (3)^2 - 6(3) + 5 = 9 - 18 + 5 = -4. Vertex is (3,βˆ’4)(3, -4).
  3. Find x-intercepts by factoring: x2βˆ’6x+5=0β€…β€ŠβŸΉβ€…β€Š(xβˆ’5)(xβˆ’1)=0x^2 - 6x + 5 = 0 \implies (x - 5)(x - 1) = 0. So, x=5x = 5 and x=1x = 1.

Explanation:

The vertex is the turning point of the parabola. We use the formula for the axis of symmetry to find the x-value and substitute it back for the y-value. Factoring the quadratic allows us to see where the curve intersects the x-axis.