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Algebra - Graphing linear and quadratic functions

Grade 10IB

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

🔑Concepts

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Linear functions follow the general form y=mx+cy = mx + c, where mm represents the gradient (slope) and cc represents the yy-intercept. The gradient determines the steepness and direction: a positive mm slopes upwards from left to right, while a negative mm slopes downwards.

Graph of a linear function with a positive gradient.
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Quadratic functions are written in the form y=ax2+bx+cy = ax^2 + bx + c. The graph is a parabola. If a>0a > 0, the parabola opens upwards (minimum point), and if a<0a < 0, it opens downwards (maximum point).

Graph of a quadratic function showing a parabola opening upwards.
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The Axis of Symmetry for a quadratic function is a vertical line passing through the vertex, given by the formula x=−b2ax = -\frac{b}{2a}. This line divides the parabola into two congruent halves.

Parabola with its vertical axis of symmetry at x=2.
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Intercepts: The yy-intercept is found by setting x=0x = 0. The xx-intercepts (or roots) are found by setting y=0y = 0. For quadratics, there can be zero, one, or two xx-intercepts depending on the discriminant Δ=b2−4ac\Delta = b^2 - 4ac.

📐Formulae

y=mx+cy = mx + c

m=y2−y1x2−x1m = \frac{y_2 - y_1}{x_2 - x_1}

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

x=−b2ax = -\frac{b}{2a}

x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

💡Examples

Problem 1:

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

Solution:

  1. Rewrite in y=mx+cy = mx + c form: 2y=4x+6  ⟹  y=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+3  ⟹  −3=2x  ⟹  x=−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=x2−6x+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=−b2a=−−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)2−6(3)+5=9−18+5=−4y = (3)^2 - 6(3) + 5 = 9 - 18 + 5 = -4. Vertex is (3,−4)(3, -4).
  3. Find x-intercepts by factoring: x2−6x+5=0  ⟹  (x−5)(x−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.

Problem 3:

Sketch the graph of the linear function y=−12x+2y = -\frac{1}{2}x + 2 and identify its intercepts.

Linear graph of y = -0.5x + 2 passing through (0,2) and (4,0).

Solution:

  1. Find the yy-intercept by setting x=0x = 0: y=−12(0)+2=2y = -\frac{1}{2}(0) + 2 = 2. So, the point is (0,2)(0, 2).
  2. Find the xx-intercept by setting y=0y = 0: 0=−12x+2⇒12x=2⇒x=40 = -\frac{1}{2}x + 2 \Rightarrow \frac{1}{2}x = 2 \Rightarrow x = 4. So, the point is (4,0)(4, 0).
  3. Plot (0,2)(0, 2) and (4,0)(4, 0) and draw a straight line through them.

Explanation:

The negative gradient m=−0.5m = -0.5 indicates that for every 2 units moved to the right, the line moves 1 unit down.

Problem 4:

Identify the vertex and intercepts for the quadratic function y=−x2+4xy = -x^2 + 4x.

Downward parabola with vertex at (2,4) and roots at 0 and 4.

Solution:

  1. For y=−x2+4xy = -x^2 + 4x, a=−1a = -1 and b=4b = 4.
  2. xx-coordinate of vertex: x=−b2a=−42(−1)=2x = -\frac{b}{2a} = -\frac{4}{2(-1)} = 2.
  3. yy-coordinate of vertex: y=−(2)2+4(2)=−4+8=4y = -(2)^2 + 4(2) = -4 + 8 = 4. Vertex is (2,4)(2, 4).
  4. xx-intercepts: set y=0y = 0: 0=−x(x−4)⇒x=00 = -x(x - 4) \Rightarrow x = 0 or x=4x = 4. Intercepts are (0,0)(0, 0) and (4,0)(4, 0).
  5. Since a<0a < 0, the parabola opens downwards.

Explanation:

The vertex (2,4)(2, 4) is the maximum point of this downward-opening parabola.