Quadratic Functions Problems with Solutions

Explore quadratic functions problems with detailed solutions and graphical interpretations.

Review: Vertex and Discriminant of Quadratic Functions

The graph of a quadratic function written as:

\[ f(x) = ax^2 + bx + c \]

has a vertex at \((h, k)\) where:

\[ h = -\frac{b}{2a} \quad \text{and} \quad k = f(h) = c - \frac{b^2}{4a} \]

The quadratic function can also be written in vertex form:

\[ f(x) = a(x - h)^2 + k \]

The discriminant \(D\) of \( ax^2 + bx + c = 0 \) is:

\[ D = b^2 - 4ac \]

  1. If \( D = 0 \), the equation has one real solution and the graph has one x-intercept.
  2. If \( D > 0 \), the equation has two real solutions and the graph has two x-intercepts.
  3. If \( D < 0 \), the equation has two complex solutions and the graph has no x-intercepts.

Problems with Detailed Solutions

Problem 1: Maximizing Profit

A company's profit (in thousands of dollars) is given by:

\[ P(x) = 5000 + 1000x - 5x^2 \]

where \(x\) is the advertising spend (in thousands of dollars).

  1. Find the advertising spend that maximizes profit.
  2. Find the maximum profit.

Solution

  1. The profit function is quadratic with \(a = -5 < 0\), so it has a maximum at the vertex:

    \[ x = h = -\frac{b}{2a} = -\frac{1000}{2(-5)} = 100 \]

    Thus, the company should spend $100,000 on advertising.

  2. The maximum profit is:

    \[ P_{\text{max}} = P(100) = 5000 + 1000(100) - 5(100)^2 = 55000 \]

    So the maximum profit is $55,000.

Graph of profit function P(x)

Problem 2: Projectile Motion

An object thrown vertically upward has its height above ground given by:

\[ S(t) = -16t^2 + v_0 t \]

where \(v_0\) is the initial velocity (ft/sec). Find \(v_0\) so the maximum height is 300 feet.

Solution

The maximum height occurs at the vertex. The vertex's \(k\)-value is:

\[ k = c - \frac{b^2}{4a} = 0 - \frac{v_0^2}{4(-16)} = \frac{v_0^2}{64} \]

Set this equal to 300:

\[ \frac{v_0^2}{64} = 300 \implies v_0^2 = 19200 \implies v_0 = \sqrt{19200} = 80\sqrt{3} \text{ ft/sec} \]

Graph of height function S(t)

Problem 3: Finding Quadratic from Intercepts

Find the quadratic function \(f\) whose graph passes through \((2, -8)\) and has x-intercepts at \((1, 0)\) and \((-2, 0)\).

Solution

Using the intercept form:

\[ f(x) = a(x - 1)(x + 2) \]

Substitute \((2, -8)\):

\[ -8 = a(2 - 1)(2 + 2) \implies -8 = 4a \implies a = -2 \]

Thus:

\[ f(x) = -2(x - 1)(x + 2) \]

Problem 4: Intersection of Quadratic and Line

Given \( f(x) = x^2 + x + 1 \) and the line \( y = mx \), find \(m\) for:

  1. 2 intersection points
  2. 1 intersection point (tangent)
  3. No intersection points

Solution

To find the points of intersection, you need to solve the system of equations \[ \left\{ \begin{array}{l} y = x^2 + x + 1 &\\ y = m x \end{array} \right. \]

Set \( mx = x^2 + x + 1 \), which simplifies to:

\[ x^2 + (1 - m)x + 1 = 0 \]

Discriminant: \( D = (1 - m)^2 - 4 \).

  1. Two intersections (\(D > 0\)): \((1 - m)^2 > 4 \implies m < -1 \text{ or } m > 3\)
  2. One intersection (\(D = 0\)): \((1 - m)^2 = 4 \implies m = -1 \text{ or } m = 3\)
  3. No intersections (\(D < 0\)): \((1 - m)^2 < 4 \implies -1 < m < 3\)

Intersection of quadratic and lines

Problem 5: Cost Minimization

The cost \(C(x) = ax^2 + bx + c\) (in thousands of dollars) of producing \(x\) items has a minimum of 120 at \(x = 2000\), and fixed cost \(C(0) = 200\). Find \(a, b, c\).

Solution

Function \( C \) is a quadratic function. Its minimum point, which is given as \( (2000,120) \) is the vertex of the graph of \( C \). Hence we can write \( C(x) \) in vertex form as follows: \[ C(x) = a(x - 2000)^2 + 120 \].

Using \(C(0) = 200\):

\[ a(0 - 2000)^2 + 120 = 200 \implies 4 \times 10^6 a = 80 \implies a = 2 \times 10^{-5} \]

Expand:

\[ C(x) = 0.00002(x - 2000)^2 + 120 = 0.00002x^2 - 0.08x + 200 \]

Thus: \( a = 0.00002,\; b = -0.08,\; c = 200 \).

Graph of cost function C(x)

Problem 6: Tangent Line to Quadratic

Find the tangent line to \( f(x) = -x^2 + x - 2 \) at \(x = 1\).

Solution

Method 1 (Algebraic):

Let tangent line be \( y = mx + b \). It passes through \((1, f(1)) = (1, -2)\):

\[ -2 = m(1) + b \implies b = -2 - m \]

For tangency, system \( y = mx + b \) and \( y = -x^2 + x - 2 \) must have exactly one solution. Substitute:

\[ mx + b = -x^2 + x - 2 \implies x^2 + (m - 1)x + (b + 2) = 0 \]

Set discriminant to zero:

\[ (m - 1)^2 - 4(1)(b + 2) = 0 \]

Substitute \( b = -2 - m \):

\[ (m - 1)^2 - 4(-m) = 0 \implies m^2 + 2m + 1 = 0 \implies (m + 1)^2 = 0 \]

Thus \( m = -1 \), \( b = -2 - (-1) = -1 \). The tangent line is:

\[ y = -x - 1 \]

Tangent line to quadratic

Method 2 (Calculus): Using derivatives (see Calculus Questions).

Practice Questions

Question 1: Find the quadratic function with x-intercepts at \((-1, 0)\) and \((3, 0)\), and y-intercept at \((0, -4)\).

Question 2: For \( f(x) = x^2 + x + c \) and line \( y = 2x \), find \(c\) for:

  1. 2 intersection points
  2. 1 intersection point
  3. No intersection points

Solutions to Practice Questions

Solution to Question 1: Using intercept form \( f(x) = a(x + 1)(x - 3) \). Substitute \((0, -4)\):

\[ -4 = a(1)(-3) \implies a = \frac{4}{3} \]

Thus: \( f(x) = \frac{4}{3}(x + 1)(x - 3) \).

Solution to Question 2: Solve \( x^2 + x + c = 2x \implies x^2 - x + c = 0 \). Discriminant: \( D = 1 - 4c \).

  1. Two intersections: \( D \gt 0 \implies c \lt \frac{1}{4} \)
  2. One intersection: \( D = 0 \implies c = \frac{1}{4} \)
  3. No intersections: \( D \lt 0 \implies c \gt \frac{1}{4} \)

Additional Resources