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Functions - Quadratic functions

Grade 12IB_AI

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

🔑Concepts

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The general form of a quadratic function is f(x)=ax2+bx+cf(x) = ax^2 + bx + c. The graph is a parabola that opens upwards if a>0a > 0 and downwards if a<0a < 0. The yy-intercept is always at (0,c)(0, c).

Graph of an upward opening parabola showing the vertex and y-intercept.
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The vertex form f(x)=a(x−h)2+kf(x) = a(x - h)^2 + k directly gives the vertex at (h,k)(h, k). The vertical line x=hx = h is the axis of symmetry, which divides the parabola into two congruent halves.

Graph of a downward parabola showing the vertical axis of symmetry passing through the vertex.
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The factored form f(x)=a(x−p)(x−q)f(x) = a(x - p)(x - q) reveals the xx-intercepts at (p,0)(p, 0) and (q,0)(q, 0). These are the roots or zeros of the quadratic equation f(x)=0f(x) = 0.

Parabola crossing the x-axis at two distinct points p and q.
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The Discriminant Δ=b2−4ac\Delta = b^2 - 4ac determines the number of xx-intercepts: if Δ>0\Delta > 0, there are two real intercepts; if Δ=0\Delta = 0, there is one (the vertex touches the axis); if Δ<0\Delta < 0, there are no real intercepts.

📐Formulae

f(x)=ax2+bx+cf(x) = ax^2 + bx + c

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 quadratic function f(x)=x2−6x+5f(x) = x^2 - 6x + 5, find the coordinates of the vertex and the xx-intercepts.

Solution:

  1. Find the xx-coordinate of the vertex: x=−b2a=−−62(1)=3x = -\frac{b}{2a} = -\frac{-6}{2(1)} = 3.
  2. Find the yy-coordinate: f(3)=(3)2−6(3)+5=9−18+5=−4f(3) = (3)^2 - 6(3) + 5 = 9 - 18 + 5 = -4. Vertex is (3,−4)(3, -4).
  3. Find xx-intercepts by setting f(x)=0f(x) = 0: x2−6x+5=0x^2 - 6x + 5 = 0 (x−1)(x−5)=0(x - 1)(x - 5) = 0 x=1,x=5x = 1, x = 5.

Explanation:

The axis of symmetry formula provides the xx-coordinate of the vertex. Substituting this back into the function gives the minimum value. Factoring the quadratic allows us to identify where the graph crosses the xx-axis.

Problem 2:

A ball is thrown into the air. Its height, hh, in meters after tt seconds is modeled by h(t)=−4.9t2+19.6t+1.4h(t) = -4.9t^2 + 19.6t + 1.4. Determine the maximum height reached by the ball.

Solution:

The maximum height occurs at the vertex since a=−4.9<0a = -4.9 < 0.

  1. Find tt at the vertex: t=−19.62(−4.9)=19.69.8=2t = -\frac{19.6}{2(-4.9)} = \frac{19.6}{9.8} = 2 seconds.
  2. Calculate h(2)h(2): h(2)=−4.9(2)2+19.6(2)+1.4h(2) = -4.9(2)^2 + 19.6(2) + 1.4 h(2)=−4.9(4)+39.2+1.4h(2) = -4.9(4) + 39.2 + 1.4 h(2)=−19.6+39.2+1.4=21h(2) = -19.6 + 39.2 + 1.4 = 21 meters.

Explanation:

In projectile motion, the vertex of the quadratic equation represents the peak of the trajectory. We find the time at which the maximum occurs and substitute it into the height function.

Problem 3:

Find the value of kk for which the equation 2x2−4x+k=02x^2 - 4x + k = 0 has exactly one real solution.

Solution:

For exactly one real solution, the discriminant must be zero: Δ=0\Delta = 0. Using a=2a = 2, b=−4b = -4, and c=kc = k: (−4)2−4(2)(k)=0(-4)^2 - 4(2)(k) = 0 16−8k=016 - 8k = 0 8k=168k = 16 k=2k = 2.

Explanation:

The discriminant Δ=b2−4ac\Delta = b^2 - 4ac determines the number of solutions. Setting it to zero ensures the parabola touches the xx-axis at exactly one point (the vertex).

Problem 4:

A rectangular garden is enclosed by 4040 meters of fencing. One side of the garden is against a straight brick wall and does not require fencing. Let xx be the length of the two sides perpendicular to the wall. Express the area AA in terms of xx and find the maximum possible area.

Diagram showing a rectangle with three sides of fencing and one side against a wall.

Solution:

Let the length perpendicular to the wall be xx. Since there are two such sides, the remaining length for the side parallel to the wall is 40−2x40 - 2x. The area A(x)=x(40−2x)=40x−2x2A(x) = x(40 - 2x) = 40x - 2x^2. This is a downward opening parabola. The maximum area occurs at the vertex. x=−b2a=−402(−2)=10x = -\frac{b}{2a} = -\frac{40}{2(-2)} = 10 m. Maximum Area A(10)=40(10)−2(10)2=400−200=200A(10) = 40(10) - 2(10)^2 = 400 - 200 = 200 m2^2.

Explanation:

We model the area as a quadratic function of xx. Finding the maximum area corresponds to finding the yy-coordinate of the vertex of the parabola.

Problem 5:

Determine the equation of the quadratic function shown in the graph, which has a vertex at (2,−4)(2, -4) and passes through the point (0,0)(0, 0).

Graph of a parabola with vertex at (2, -4) passing through the origin.

Solution:

Use the vertex form f(x)=a(x−h)2+kf(x) = a(x - h)^2 + k. Substitute the vertex (2,−4)(2, -4): f(x)=a(x−2)2−4f(x) = a(x - 2)^2 - 4. To find aa, substitute the point (0,0)(0, 0): 0=a(0−2)2−40 = a(0 - 2)^2 - 4 0=4a−40 = 4a - 4 4a=4  ⟹  a=14a = 4 \implies a = 1. The equation is f(x)=(x−2)2−4f(x) = (x - 2)^2 - 4, or f(x)=x2−4xf(x) = x^2 - 4x.

Explanation:

Vertex form is the most efficient starting point when the coordinates of the turning point are known. Solving for 'a' ensures the parabola passes through the secondary given point.