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Algebra - Transformation of quadratic functions-extended

Grade 10IB

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

🔑Concepts

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The basic parent function for quadratics is f(x)=x2f(x) = x^2. This is a parabola with its vertex at (0,0)(0, 0) and an axis of symmetry at x=0x = 0. Every transformation is defined relative to this starting shape.

Graph of the parent function y = x^2
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Horizontal and vertical translations are shifts in the position of the vertex. In vertex form f(x)=a(x−h)2+kf(x) = a(x - h)^2 + k, the value hh shifts the graph left (h<0h < 0) or right (h>0h > 0), and kk shifts it down (k<0k < 0) or up (k>0k > 0).

Parabola shifted 2 units right and 1 unit up
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Vertical stretching and shrinking are controlled by the coefficient aa. If ∣a∣>1|a| > 1, the parabola is vertically stretched (it looks narrower). If 0<∣a∣<10 < |a| < 1, it is vertically compressed (it looks wider). If a<0a < 0, the parabola is reflected across the xx-axis.

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The vertex form f(x)=a(x−h)2+kf(x) = a(x - h)^2 + k is the most useful form for identifying transformations immediately. The point (h,k)(h, k) is the vertex, and the vertical line x=hx = h is the axis of symmetry.

📐Formulae

f(x)=a(x−h)2+kf(x) = a(x - h)^2 + k

Vertex=(h,k)\text{Vertex} = (h, k)

Axis of Symmetry: x=h\text{Axis of Symmetry: } x = h

h=−b2ah = -\frac{b}{2a}

k=c−b24ak = c - \frac{b^2}{4a}

💡Examples

Problem 1:

Describe the transformations required to map f(x)=x2f(x) = x^2 onto g(x)=−3(x+4)2−2g(x) = -3(x + 4)^2 - 2.

Solution:

g(x)=−3(x−(−4))2−2g(x) = -3(x - (-4))^2 - 2.

  1. Horizontal translation: 4 units to the left (h=−4h = -4).
  2. Vertical stretch: Factor of 3 (∣a∣=3|a| = 3).
  3. Reflection: Reflected across the xx-axis (aa is negative).
  4. Vertical translation: 2 units downwards (k=−2k = -2).

Explanation:

Compare the given equation to the vertex form a(x−h)2+ka(x - h)^2 + k. Identify a=−3a = -3, h=−4h = -4, and k=−2k = -2.

Problem 2:

Convert the quadratic function f(x)=2x2−12x+11f(x) = 2x^2 - 12x + 11 into vertex form.

Solution:

f(x)=2(x2−6x)+11f(x) = 2(x^2 - 6x) + 11 f(x)=2(x2−6x+9−9)+11f(x) = 2(x^2 - 6x + 9 - 9) + 11 f(x)=2((x−3)2−9)+11f(x) = 2((x - 3)^2 - 9) + 11 f(x)=2(x−3)2−18+11f(x) = 2(x - 3)^2 - 18 + 11 f(x)=2(x−3)2−7f(x) = 2(x - 3)^2 - 7

Explanation:

Factor out the coefficient of x2x^2 from the first two terms. Complete the square inside the bracket by adding and subtracting (b2)2(\frac{b}{2})^2, which is (−62)2=9(\frac{-6}{2})^2 = 9. Simplify to reach the vertex form.

Problem 3:

Find the equation of a parabola that has been reflected across the xx-axis, stretched vertically by a factor of 2, shifted 5 units right, and 3 units up.

Solution:

a=−2a = -2 h=5h = 5 k=3k = 3 Substituting into y=a(x−h)2+ky = a(x - h)^2 + k: y=−2(x−5)2+3y = -2(x - 5)^2 + 3

Explanation:

The reflection makes aa negative. The stretch factor 2 sets ∣a∣=2|a| = 2. Right shift sets hh as positive 5. Upward shift sets kk as positive 3.

Problem 4:

Identify the function g(x)g(x) that results from reflecting f(x)=x2f(x) = x^2 over the xx-axis, then shifting it 33 units to the left and 44 units down. Sketch the result.

Graph of g(x) = -(x+3)^2 - 4

Solution:

  1. Reflection over the xx-axis: Multiply by −1-1, so y=−x2y = -x^2.
  2. Shift 33 units left: Replace xx with x+3x + 3, so y=−(x+3)2y = -(x + 3)^2.
  3. Shift 44 units down: Subtract 44, so g(x)=−(x+3)2−4g(x) = -(x + 3)^2 - 4.

Explanation:

Starting from the origin, the vertex (0,0)(0,0) moves to (−3,−4)(-3, -4). The negative sign in front of the bracket indicates the parabola opens downwards.

Problem 5:

A quadratic function has its vertex at (1,2)(1, 2) and passes through the point (2,5)(2, 5). Find the equation of the function in vertex form.

Parabola with vertex (1,2) passing through (2,5)

Solution:

  1. Use the vertex form y=a(x−h)2+ky = a(x - h)^2 + k with h=1h = 1 and k=2k = 2: y=a(x−1)2+2y = a(x - 1)^2 + 2.
  2. Substitute the point (2,5)(2, 5) into the equation to find aa: 5=a(2−1)2+25 = a(2 - 1)^2 + 2 5=a(1)2+25 = a(1)^2 + 2 5=a+25 = a + 2 a=3a = 3
  3. The final equation is y=3(x−1)2+2y = 3(x - 1)^2 + 2.

Explanation:

The vertex determines the horizontal and vertical shifts. The second point allows us to solve for the vertical stretch factor aa.

Transformation of quadratic functions-extended Grade 10 Notes & Examples