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Functions - Quadratic Functions

Grade 11IB

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

🔑Concepts

General Form f(x)=ax2+bx+cf(x) = ax^2 + bx + c: The graph of a quadratic function is a symmetrical curve called a parabola. If the coefficient aa is positive (a>0a > 0), the parabola opens upwards like a 'U' shape, creating a minimum point. If aa is negative (a<0a < 0), the parabola opens downwards like an inverted 'U', creating a maximum point. The constant cc is the yy-intercept, representing the point (0,c)(0, c) where the curve crosses the vertical axis.

Vertex Form f(x)=a(xh)2+kf(x) = a(x - h)^2 + k: This form is particularly useful for identifying the vertex (h,k)(h, k), which is the turning point of the parabola. Visually, the graph is a transformation of the parent function y=x2y = x^2, shifted horizontally by hh units and vertically by kk units. The vertical line x=hx = h acts as the axis of symmetry, splitting the parabola into two identical mirror images.

Factored (Intercept) Form f(x)=a(xp)(xq)f(x) = a(x - p)(x - q): This form highlights the xx-intercepts, also known as the roots or zeros of the function, which occur at (p,0)(p, 0) and (q,0)(q, 0). On a graph, these are the points where the curve crosses the horizontal xx-axis. The axis of symmetry is always located exactly midway between these two points at x=p+q2x = \frac{p + q}{2}.

The Discriminant Δ=b24ac\Delta = b^2 - 4ac: The discriminant determines the nature of the roots and how the graph interacts with the xx-axis. If Δ>0\Delta > 0, the parabola crosses the xx-axis at two distinct points. If Δ=0\Delta = 0, the parabola's vertex touches the xx-axis at exactly one point (a repeated root). If Δ<0\Delta < 0, the entire parabola floats above or below the xx-axis and never touches it, meaning there are no real roots.

Axis of Symmetry and Turning Point: Every parabola has a vertical line of symmetry with the equation x=b2ax = -\frac{b}{2a}. The vertex always lies on this line. To find the yy-coordinate of the vertex (the maximum or minimum value), substitute x=b2ax = -\frac{b}{2a} back into the original function. Visually, this turning point is the peak of a hill (a<0a < 0) or the bottom of a valley (a>0a > 0).

Vertical Dilations and Reflections: The value of aa controls the 'steepness' of the curve. As a|a| increases (a>1|a| > 1), the parabola undergoes a vertical stretch, making it look thinner or narrower. When 0<a<10 < |a| < 1, the parabola undergoes a vertical compression, making it appear wider or flatter. A negative sign in front of aa reflects the graph across the xx-axis.

📐Formulae

Standard Form: y=ax2+bx+cy = ax^2 + bx + c

Vertex Form: y=a(xh)2+ky = a(x - h)^2 + k

Factored Form: y=a(xp)(xq)y = a(x - p)(x - q) where p,qp, q are roots

Quadratic Formula: x=b±b24ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Discriminant: Δ=b24ac\Delta = b^2 - 4ac

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

Vertex Coordinates: (h,k)=(b2a,f(b2a))(h, k) = (-\frac{b}{2a}, f(-\frac{b}{2a}))

💡Examples

Problem 1:

Given the quadratic function f(x)=2x212x+13f(x) = 2x^2 - 12x + 13, convert it into vertex form and state the coordinates of the vertex and the equation of the axis of symmetry.

Solution:

  1. Factor out the aa coefficient from the first two terms: f(x)=2(x26x)+13f(x) = 2(x^2 - 6x) + 13
  2. Complete the square inside the parentheses. Take half of 6-6, which is 3-3, and square it to get 99: f(x)=2(x26x+99)+13f(x) = 2(x^2 - 6x + 9 - 9) + 13
  3. Distribute the 22 to the 9-9 to move it out of the parentheses: f(x)=2(x26x+9)18+13f(x) = 2(x^2 - 6x + 9) - 18 + 13
  4. Simplify: f(x)=2(x3)25f(x) = 2(x - 3)^2 - 5
  5. The vertex is (h,k)=(3,5)(h, k) = (3, -5). The axis of symmetry is x=3x = 3.

Explanation:

By completing the square, we transformed the standard form into vertex form a(xh)2+ka(x-h)^2+k. This allows us to read the vertex directly and identify the axis of symmetry from the hh value.

Problem 2:

Find the range of values for kk such that the equation x2kx+9=0x^2 - kx + 9 = 0 has two distinct real roots.

Solution:

  1. For two distinct real roots, the discriminant must be greater than zero: Δ>0\Delta > 0
  2. Identify coefficients: a=1,b=k,c=9a = 1, b = -k, c = 9
  3. Calculate Δ\Delta: b24ac=(k)24(1)(9)=k236b^2 - 4ac = (-k)^2 - 4(1)(9) = k^2 - 36
  4. Set the inequality: k236>0k^2 - 36 > 0
  5. Factor the difference of squares: (k6)(k+6)>0(k - 6)(k + 6) > 0
  6. Solve the quadratic inequality: k<6k < -6 or k>6k > 6.

Explanation:

The discriminant determines the number of real roots. To ensure two distinct roots, we solve for when the discriminant is strictly positive, using a sign diagram or inequality rules.