Lesson 5.1: Quadratic Functions

New Concept

A quadratic function, like $f(x) = ax^2 + bx + c$, models a U-shaped curve called a parabola. We'll explore its key features, find its vertex (maximum/minimum value), and use these skills to solve real-world problems.

What’s next

You're just getting started! Next, you'll work through interactive examples to identify a parabola's vertex and intercepts, and then apply these skills in challenge problems.

Property

The graph of a quadratic function is a U-shaped curve called a parabola. One important feature of the graph is that it has an extreme point, called the vertex.

The graph is also symmetric with a vertical line drawn through the vertex, called the axis of symmetry.
The y-intercept is the point at which the parabola crosses the $y$-axis.
The x-intercepts are the points at which the parabola crosses the $x$-axis.
If they exist, the $x$-intercepts represent the zeros, or roots, of the quadratic function.

Examples

• A parabola has its vertex at $(-2, 5)$ and opens upward. Its axis of symmetry is the line $x = -2$, and its minimum value is $5$. It crosses the y-axis at $(0, 9)$, which is its y-intercept.
• For a parabola that opens downward with a vertex at $(4, -1)$, the axis of symmetry is $x=4$ and the maximum value of the function is $-1$.
• If a quadratic function has zeros at $x=-5$ and $x=1$, its graph crosses the x-axis at the points $(-5, 0)$ and $(1, 0)$.

Property

A quadratic function is a polynomial function of degree two. The graph of a quadratic function is a parabola.
The general form of a quadratic function is $f(x) = ax^2 + bx + c$ where $a$, $b$, and $c$ are real numbers and $a \neq 0$.
The standard form of a quadratic function is $f(x) = a(x - h)^2 + k$ where $a \neq 0$. This is also known as the vertex form. The vertex $(h, k)$ is located at

Examples

• The function $f(x) = 3x^2 - 6x + 2$ is in general form with $a=3$, $b=-6$, and $c=2$. Since $a>0$, the parabola opens upward.
• The function $g(x) = -4(x - 1)^2 + 5$ is in standard form. Its vertex is at $(1, 5)$, and because $a=-4$ is negative, the parabola opens downward.
• To convert $f(x) = 2(x-3)^2-8$ to general form, expand it: $f(x) = 2(x^2-6x+9)-8 = 2x^2-12x+18-8 = 2x^2-12x+10$.

Explanation

General form, $f(x) = ax^2+bx+c$, is useful for finding the y-intercept, which is $c$. Standard form, $f(x) = a(x-h)^2+k$, is called vertex form because it directly shows you the vertex coordinates, $(h, k)$.
If $a > 0$, the parabola opens upward. If $a < 0$, the parabola opens downward.

Property

Given a quadratic function in general form, $f(x) = ax^2 + bx + c$, the vertex of the parabola can be found.

1. Identify the coefficients $a$, $b$, and $c$.
2. Find $h$, the $x$-coordinate of the vertex, by substituting $a$ and $b$ into the formula $h = -\dfrac{b}{2a}$.
3. Find $k$, the $y$-coordinate of the vertex, by evaluating $k = f(h)$.

Examples

• For $f(x) = x^2 - 8x + 1$, find the vertex. Here, $a=1, b=-8$. The x-coordinate is $h = -\frac{-8}{2(1)} = 4$. The y-coordinate is $k = f(4) = 4^2 - 8(4) + 1 = -15$. The vertex is $(4, -15)$.
• Find the vertex of $f(x) = -2x^2 - 4x + 5$. The x-coordinate is $h = -\frac{-4}{2(-2)} = -1$. The y-coordinate is $k = f(-1) = -2(-1)^2 - 4(-1) + 5 = 7$. The vertex is $(-1, 7)$.
• To rewrite $f(x) = 3x^2 + 18x + 30$ in standard form, first find the vertex. $h = -\frac{18}{2(3)} = -3$. $k = 3(-3)^2 + 18(-3) + 30 = 3$. The standard form is $f(x) = 3(x+3)^2+3$.

Explanation

The vertex formula $h = -\frac{b}{2a}$ finds the x-coordinate that lies on the axis of symmetry. Plugging this x-value back into the function gives you the corresponding y-value, which is the function's maximum or minimum output.

Property

The domain of any quadratic function is all real numbers unless the context of the function presents some restrictions.
The range of a quadratic function written in general form $f(x) = ax^2 + bx + c$ with a positive $a$ value is $f(x) \geq f(-\frac{b}{2a})$, or $[f(-\frac{b}{2a}), \infty)$.
The range of a quadratic function written in general form with a negative $a$ value is $f(x) \leq f(-\frac{b}{2a})$, or $(-\infty, f(-\frac{b}{2a})]$.
For a function in standard form $f(x) = a(x-h)^2+k$, the range is $[k, \infty)$ if $a>0$ and $(-\infty, k]$ if $a<0$.

Examples

• The function $f(x) = (x-5)^2 + 3$ has a vertex at $(5, 3)$ and opens up ($a=1>0$). The domain is all real numbers, $(-\infty, \infty)$, and the range is $[3, \infty)$.
• For $f(x) = -2x^2 + 8x - 1$, first find the vertex. $h = -\frac{8}{2(-2)} = 2$. $k = f(2) = -2(2)^2+8(2)-1 = 7$. Since $a<0$, the range is $(-\infty, 7]$.
• A parabola has a maximum value of 10. Its domain is all real numbers, $(-\infty, \infty)$, and its range is $(-\infty, 10]$.

Explanation

The domain of a quadratic function is always all real numbers because you can plug any x-value into it. The range is limited by the vertex. If the parabola opens up, the range includes all y-values from the vertex up.

Property

Given a quadratic function $f(x)$, find the $y$- and $x$-intercepts.

1. Evaluate $f(0)$ to find the $y$-intercept. The $y$-intercept is the point where the graph crosses the $y$-axis.
2. Solve the quadratic equation $f(x) = 0$ to find the $x$-intercepts. The $x$-intercepts are the points where the graph crosses the $x$-axis. This can be done by factoring, using the quadratic formula, or rewriting in standard form.

Examples

• Find the intercepts of $f(x) = x^2 - 7x + 10$. The y-intercept is $f(0)=10$, so at $(0, 10)$. For x-intercepts, solve $x^2 - 7x + 10 = 0$, which factors to $(x-2)(x-5)=0$. The x-intercepts are at $(2, 0)$ and $(5, 0)$.
• Find the y-intercept of $g(x) = 3(x+2)^2-5$. Evaluate $g(0) = 3(0+2)^2-5 = 3(4)-5 = 7$. The y-intercept is $(0, 7)$.
• Find the x-intercepts of $f(x) = (x-4)^2 - 9$. Set $f(x)=0$ to get $(x-4)^2 = 9$. Taking the square root, $x-4 = \pm 3$, so $x = 4 \pm 3$. The x-intercepts are at $(7, 0)$ and $(1, 0)$.

Explanation

Intercepts are where the parabola crosses the axes. Find the y-intercept by setting $x=0$. Find the x-intercepts, or zeros, by setting the whole function equal to zero and solving for $x$. A parabola can have 0, 1, or 2 x-intercepts.

Property

The output of the quadratic function at the vertex is the maximum or minimum value of the function, depending on the orientation of the parabola. The vertex is $(h,k)$ where $h = -\frac{b}{2a}$ and $k=f(h)$.
If $a > 0$, the parabola opens upward and has a minimum value of $k$.
If $a < 0$, the parabola opens downward and has a maximum value of $k$.

Examples

• A ball's height is given by $H(t) = -16t^2 + 64t + 5$. To find the maximum height, find the vertex. The time to max height is $t = -\frac{64}{2(-16)} = 2$ seconds. The max height is $H(2) = -16(2)^2+64(2)+5 = 69$ feet.
• A farmer wants to enclose a rectangular field with 100 feet of fence to maximize area. The area is $A(L) = L(50-L) = -L^2+50L$. The length for max area is $L = -\frac{50}{2(-1)} = 25$ feet. The dimensions should be 25 by 25 feet.
• The profit from selling shoes is $P(x) = -0.5x^2 + 40x - 150$. To maximize profit, find the vertex. $x = -\frac{40}{2(-0.5)} = 40$. Selling 40 pairs of shoes will yield the maximum profit.

Explanation

The vertex's y-coordinate, $k$, represents the highest or lowest value a quadratic function can achieve. This is incredibly useful for real-world problems like finding a maximum profit, maximum height, or minimum cost.To find the maximum revenue from a revenue function, find the vertex of the quadratic equation.