Quadratic Functions Problems with Solutions

6 real-world word problems — profit, projectile motion, cost, tangent lines — with full step-by-step solutions

Overview

Work through six quadratic function word problems with detailed solutions and graphical interpretations. Topics include profit maximization, projectile motion, cost minimization, tangent lines, and quadratic–line intersections.

Review: Vertex and Discriminant

For \( f(x) = ax^2 + bx + c \), the vertex \( (h, k) \) is:

\[ h = -\frac{b}{2a}, \qquad k = f(h) = c - \frac{b^2}{4a} \]

Vertex form:

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

The discriminant \( D = b^2 - 4ac \) determines the number of x-intercepts of \( ax^2 + bx + c = 0 \):

Discriminant Solutions x-intercepts
\( D > 0 \)Two distinct realTwo
\( D = 0 \)One real (double root)One (vertex touches x-axis)
\( D < 0 \)Two complex conjugatesNone

Problems with Detailed Solutions

Problem 1 — Maximizing profit

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

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

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

  1. What advertising spend maximizes profit?
  2. What is the maximum profit?
Show solution

  1. The profit function is a downward parabola (\( a = -5 < 0 \)), so the maximum is at the vertex.

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

    So the company should spend \( x = 100 \) thousand dollars, i.e. $100,000, on advertising.

  2. \[ P(100) = 5000 + 1000(100) - 5(100)^2 = 5000 + 100{,}000 - 50{,}000 = 55{,}000 \]

    The result \( 55{,}000 \) is in thousands of dollars, so the maximum profit is $55,000,000 (55 million dollars).

Profit function P(x) = 5000 + 1000x − 5x² showing a maximum at x = 100 (thousands of dollars)
Figure 1 — \( P(x) = 5000 + 1000x - 5x^2 \), with maximum at \( x = 100 \).

Problem 2 — Projectile motion

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

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

where \( v_0 \) is the initial velocity in ft/sec. What initial velocity \( v_0 \) is needed to reach a maximum height of 300 feet?

Show solution

The maximum height is the \(k\)-value of the vertex (with \(a = -16\), \(b = v_0\), \(c = 0\)):

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

Set \( k = 300 \):

\[ \frac{v_0^2}{64} = 300 \;\Rightarrow\; v_0^2 = 19200 \;\Rightarrow\; v_0 = \sqrt{19200} = 80\sqrt{3} \approx 138.6 \text{ ft/sec} \]
Height function S(t) = −16t² + v₀t for projectile motion, showing maximum height at the vertex
Figure 2 — Height function \( S(t) = -16t^2 + v_0 t \).

Problem 3 — Quadratic from x-intercepts and one point

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

Show solution

Use the intercept form with \( r_1 = 1 \) and \( r_2 = -2 \):

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

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

\[ -8 = a(2 - 1)(2 + 2) \;\Rightarrow\; -8 = 4a \;\Rightarrow\; a = -2 \] \[ f(x) = -2(x - 1)(x + 2) \]

Check: \( f(2) = -2(1)(4) = -8 \) ✓; \( f(1) = 0 \) ✓; \( f(-2) = 0 \) ✓

Expanded: \( f(x) = -2x^2 - 2x + 4 \).

Problem 4 — Intersections of a parabola and a line

For the parabola \( f(x) = x^2 + x + 1 \) and the line \( y = mx \), find the values of \(m\) that give:

  1. 2 intersection points
  2. 1 intersection point (tangent)
  3. no intersection points
Show solution

Set the equations equal:

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

Discriminant:

\[ D = (1 - m)^2 - 4 \]

  1. Two intersections (\( D > 0 \)):

    \[ (1 - m)^2 > 4 \;\Rightarrow\; 1 - m > 2 \;\text{ or }\; 1 - m < -2 \;\Rightarrow\; m < -1 \;\text{ or }\; m > 3 \]

  2. One intersection (\( D = 0 \)):

    \[ (1 - m)^2 = 4 \;\Rightarrow\; m = -1 \;\text{ or }\; m = 3 \]

  3. No intersections (\( D < 0 \)):

    \[ (1 - m)^2 < 4 \;\Rightarrow\; -1 < m < 3 \]
Parabola y = x² + x + 1 with several lines y = mx, showing two-intersection, tangent, and no-intersection cases
Figure 3 — Intersections of \( y = x^2 + x + 1 \) with lines \( y = mx \).

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 \(a\), \(b\), and \(c\).

Show solution

Vertex is \( (2000, 120) \), so vertex form is:

\[ C(x) = a(x - 2000)^2 + 120 \]

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

\[ 200 = a(0 - 2000)^2 + 120 \;\Rightarrow\; 80 = 4{,}000{,}000a \;\Rightarrow\; a = 0.00002 \]

Expand:

\[ C(x) = 0.00002(x^2 - 4000x + 4{,}000{,}000) + 120 = 0.00002x^2 - 0.08x + 200 \]

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

Check: \( C(2000) = 0.00002(4{,}000{,}000) - 0.08(2000) + 200 = 80 - 160 + 200 = 120 \) ✓

Graph of the cost function C(x) = 0.00002x² − 0.08x + 200, showing a minimum at x = 2000 and y-intercept at 200
Figure 4 — Cost function \( C(x) = 0.00002x^2 - 0.08x + 200 \), minimum at \( (2000, 120) \).

Problem 6 — Tangent line to a parabola

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

Show solution

Method 1 — Algebraic (using the discriminant):

First find the point of tangency:

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

Let the tangent be \( y = mx + b \) through \( (1, -2) \), so \( b = -2 - m \).

For tangency, the system must have exactly one solution. Set equal:

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

Require \( D = 0 \):

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

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

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

Then \( b = -2 - (-1) = -1 \), so the tangent is:

\[ y = -x - 1 \]

Method 2 — Calculus: the slope at \( x = 1 \) is \( f'(1) = -2(1) + 1 = -1 \). Through \( (1, -2) \): \( y + 2 = -1(x - 1) \Rightarrow y = -x - 1 \) ✓

Parabola y = −x² + x − 2 with its tangent line y = −x − 1 at the point (1, −2)
Figure 5 — Tangent to \( y = -x^2 + x - 2 \) at \( (1, -2) \).

Practice Questions

Practice 1 — Quadratic from x-intercepts and y-intercept

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

Show solution

Intercept form with \( r_1 = -1 \), \( r_2 = 3 \):

\[ f(x) = a(x + 1)(x - 3) \]

Use \( f(0) = -4 \):

\[ -4 = a(1)(-3) \;\Rightarrow\; a = \frac{4}{3} \] \[ f(x) = \frac{4}{3}(x + 1)(x - 3) = \frac{4}{3}x^2 - \frac{8}{3}x - 4 \]

Practice 2 — Intersections of a parabola and a line

For the parabola \( f(x) = x^2 + x + c \) and line \( y = 2x \), find the values of \(c\) that give:

  1. 2 intersection points
  2. 1 intersection point
  3. no intersection points
Show solution

Set equal:

\[ x^2 + x + c = 2x \;\Rightarrow\; x^2 - x + c = 0 \]

Discriminant:

\[ D = 1 - 4c \]
  1. Two intersections: \( D > 0 \) → \( c < \tfrac{1}{4} \)
  2. One intersection: \( D = 0 \) → \( c = \tfrac{1}{4} \)
  3. No intersections: \( D < 0 \) → \( c > \tfrac{1}{4} \)

Frequently Asked Questions

How do I find the maximum or minimum of a quadratic function?

For \(f(x) = ax^2 + bx + c\), the maximum or minimum is at the vertex \( (h, k) \) where \(h = -b/(2a)\) and \(k = c - b^2/(4a)\). If \(a > 0\) the vertex is a minimum; if \(a < 0\) it is a maximum.

How does the discriminant tell me the number of x-intercepts?

The discriminant is \(D = b^2 - 4ac\). If \(D > 0\), the parabola has two distinct x-intercepts; if \(D = 0\), exactly one (the vertex touches the x-axis); if \(D < 0\), no real x-intercepts.

How do I find a quadratic from its x-intercepts and one other point?

Use the intercept form \(f(x) = a(x - r_1)(x - r_2)\) with the given x-intercepts \(r_1\) and \(r_2\). Substitute the other point to solve for \(a\), then expand into standard form if needed.

How do I find the tangent line to a parabola at a given point?

Two methods work. Algebra: write the tangent as \(y = mx + b\) through the point, set it equal to the parabola, and require the resulting quadratic to have discriminant zero. Calculus: the slope of the tangent is \(f'(x)\); evaluate at the point to get \(m\).

Additional Resources

Last updated: October 20, 2026