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Functions - Quadratics (Quadratic functions)

Grade 11IB_AA

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

🔑Concepts

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The vertex (h,k)(h, k) of a quadratic function f(x)=ax2+bx+cf(x) = ax^2 + bx + c represents the maximum or minimum point. The vertical line x=hx = h (where h=−b2ah = -\frac{b}{2a}) is the axis of symmetry, dividing the parabola into two mirror-image halves.

Parabola showing vertex at (2,1) and axis of symmetry x=2
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The discriminant Δ=b2−4ac\Delta = b^2 - 4ac determines the number of xx-intercepts (roots). If Δ>0\Delta > 0, there are two distinct real roots; if Δ=0\Delta = 0, there is one repeated real root (the vertex touches the xx-axis); if Δ<0\Delta < 0, there are no real roots.

Graph showing two parabolas, one with two x-intercepts and one with none.
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The sign of the leading coefficient aa determines the concavity. If a>0a > 0, the parabola opens upwards (concave up) and has a minimum. If a<0a < 0, the parabola opens downwards (concave down) and has a maximum.

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The yy-intercept of the function is found at (0,c)(0, c) in the standard form f(x)=ax2+bx+cf(x) = ax^2 + bx + c.

📐Formulae

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

Δ=b2−4ac\Delta = b^2 - 4ac

x=−b2ax = -\frac{b}{2a}

f(x)=a(x−h)2+kf(x) = a(x-h)^2 + k

f(x)=a(x−p)(x−q)f(x) = a(x-p)(x-q)

💡Examples

Problem 1:

Given the function f(x)=2x2−12x+10f(x) = 2x^2 - 12x + 10, express the function in vertex form and state the coordinates of the vertex.

Solution:

f(x)=2(x2−6x)+10f(x) = 2(x^2 - 6x) + 10 f(x)=2(x2−6x+9−9)+10f(x) = 2(x^2 - 6x + 9 - 9) + 10 f(x)=2((x−3)2−9)+10f(x) = 2((x - 3)^2 - 9) + 10 f(x)=2(x−3)2−18+10f(x) = 2(x - 3)^2 - 18 + 10 f(x)=2(x−3)2−8f(x) = 2(x - 3)^2 - 8 The vertex is (3,−8)(3, -8).

Explanation:

To convert to vertex form, we complete the square. Factor out the leading coefficient a=2a=2 from the xx terms, then add and subtract (b2)2(\frac{b}{2})^2 inside the parentheses.

Problem 2:

Find the range of values for kk such that the equation x2−5x+k=0x^2 - 5x + k = 0 has no real roots.

Solution:

For no real roots, the discriminant must be less than zero: Δ<0\Delta < 0 b2−4ac<0b^2 - 4ac < 0 (−5)2−4(1)(k)<0(-5)^2 - 4(1)(k) < 0 25−4k<025 - 4k < 0 25<4k25 < 4k k>254k > \frac{25}{4}

Explanation:

The nature of the roots is determined by the discriminant Δ\Delta. No real roots occur when the value inside the square root of the quadratic formula is negative.

Problem 3:

A parabola has xx-intercepts at −2-2 and 44, and passes through the point (0,16)(0, 16). Find the equation of the function in the form f(x)=ax2+bx+cf(x) = ax^2 + bx + c.

Solution:

Using factored form f(x)=a(x−p)(x−q)f(x) = a(x - p)(x - q): f(x)=a(x−(−2))(x−4)f(x) = a(x - (-2))(x - 4) f(x)=a(x+2)(x−4)f(x) = a(x + 2)(x - 4) Substitute (0,16)(0, 16): 16=a(0+2)(0−4)16 = a(0 + 2)(0 - 4) 16=a(2)(−4)16 = a(2)(-4) 16=−8a⇒a=−216 = -8a \Rightarrow a = -2 Expand to general form: f(x)=−2(x2−2x−8)f(x) = -2(x^2 - 2x - 8) f(x)=−2x2+4x+16f(x) = -2x^2 + 4x + 16

Explanation:

Start with the factored form since the roots are given. Substitute the third known point to find the leading coefficient aa, then expand the brackets to reach the general form.

Problem 4:

Identify the coordinates of the xx-intercepts and the vertex for the function f(x)=−(x−1)(x−5)f(x) = -(x-1)(x-5). Sketch the graph.

Graph of y = -(x-1)(x-5) with intercepts at 1 and 5 and vertex at (3,4).

Solution:

  1. The xx-intercepts are found by setting f(x)=0f(x) = 0. From −(x−1)(x−5)=0-(x-1)(x-5) = 0, we get x=1x = 1 and x=5x = 5.
  2. The xx-coordinate of the vertex (hh) is the midpoint of the intercepts: h=1+52=3h = \frac{1+5}{2} = 3.
  3. Substitute h=3h = 3 into the function to find the yy-coordinate (kk): k=f(3)=−(3−1)(3−5)=−(2)(−2)=4k = f(3) = -(3-1)(3-5) = -(2)(-2) = 4.
  4. The vertex is (3,4)(3, 4). Since a=−1a = -1, the parabola opens downwards.

Explanation:

Using the factored form f(x)=a(x−p)(x−q)f(x) = a(x-p)(x-q) allows us to immediately identify the roots pp and qq. The axis of symmetry always lies exactly halfway between these roots.

Problem 5:

Determine the equation of the quadratic function shown in the diagram, which has a vertex at (0,−3)(0, -3) and passes through the point (2,5)(2, 5).

Graph of a parabola with vertex at (0,-3) passing through (2,5).

Solution:

  1. Start with the vertex form f(x)=a(x−h)2+kf(x) = a(x-h)^2 + k. Given vertex (0,−3)(0, -3), h=0h = 0 and k=−3k = -3.
  2. The equation is f(x)=a(x−0)2−3f(x) = a(x-0)^2 - 3, or f(x)=ax2−3f(x) = ax^2 - 3.
  3. Use the point (2,5)(2, 5) to solve for aa: 5=a(2)2−35 = a(2)^2 - 3.
  4. 5=4a−3  ⟹  8=4a  ⟹  a=25 = 4a - 3 \implies 8 = 4a \implies a = 2.
  5. The final equation is f(x)=2x2−3f(x) = 2x^2 - 3.

Explanation:

When the vertex is on the yy-axis, the function simplifies to the form ax2+cax^2 + c. We use a known point on the curve to calculate the vertical stretch factor aa.