07/14 7. Quadratic Functions and Equations

Recognizing Quadratic Functions

A quadratic expression is a polynomial of degree $2$. A quadratic equation can be written as

A quadratic function is written as

The coefficient $a$ controls the basic shape:

• If $a>0$, the graph opens upward and has a minimum.

• If $a<0$, the graph opens downward and has a maximum.

• A larger value of $\lvert a\rvert$ produces a narrower graph.

• A smaller value of $\lvert a\rvert$ produces a wider graph.

In standard form, the constant term gives the $y$-intercept, $(0,c)$, because $f(0)=c$.

The three most useful forms are:

• Standard form: $f(x)=ax^2+bx+c$

• Vertex form: $f(x)=a(x-h)^2+k$, with vertex $(h,k)$

• Factored form: f(x)=a(x-r_1)(x-r_2)\, with zeros $x=r_1$ and $x=r_2$

For example, $f(x)=2(x-3)^2-5$ has vertex $(3,-5)$, axis of symmetry $x=3$, and opens upward. The function $f(x)=(x-2)(x+4)$ has zeros $x=2$ and $x=-4$.

Takeaway: Identify the form first. Standard form highlights the coefficients and intercept, vertex form highlights the vertex, and factored form highlights the zeros.

Graphing Parabolas

The key features of a parabola are its vertex, axis of symmetry, intercepts, and direction of opening. For a function in standard form,

the axis of symmetry is

Substitute this $x$-value into the function to find the vertex's $y$-coordinate.

Example

For

identify $a=1$, $b=-6$, and $c=5$. Then

Evaluate the function:

The vertex is $(3,-4)$, and the parabola opens upward because $a>0$. The $y$-intercept is $(0,5)$.

To find the $x$-intercepts, factor:

Thus, the $x$-intercepts are $(1,0)$ and $(5,0)$. They are symmetric around $x=3$.

Graphing sequence

1. Determine the direction of opening from the sign of $a$.

2. Find the vertex using $x=-\frac{b}{2a}$.

3. Draw the axis of symmetry.

4. Find the $x$- and $y$-intercepts when useful.

5. Plot additional points and connect them with a smooth curve.

Takeaway: The vertex and axis of symmetry organize the graph, while the intercepts help locate where the parabola crosses the coordinate axes.

Solving by Factoring

Factoring is often the fastest way to solve a quadratic when the expression factors easily. The zero-product property allows a product equal to zero to be separated into individual equations.

Example with leading coefficient $1$

Solve

Factor:

Set each factor equal to zero:

Therefore,

Example with a different leading coefficient

Solve

Factor by grouping:

Then

so

The solutions are

Always check that a factored equation is equivalent to the original equation and that both solutions have been included.

Takeaway: Factoring transforms one quadratic equation into simpler linear equations, but it is not available or convenient for every quadratic.

Completing the Square

Completing the square creates a perfect-square binomial. The identity

shows why adding the square of half the linear coefficient works. For an expression of the form $x^2+bx$, add

Example

Solve

Move the constant term:

Half of $6$ is $3$, and $3^2=9$. Add $9$ to both sides:

Rewrite the left side:

Take both square roots:

Therefore,

Completing the square is also useful for converting a function into vertex form. If the coefficient of $x^2$ is not $1$, divide the equation by that coefficient first when doing so is convenient.

Takeaway: Completing the square is especially useful for revealing the vertex and solving equations that do not factor easily.

The Quadratic Formula and Method Selection

The quadratic formula solves every quadratic equation in standard form:

Its formula is

The expression under the square root is the discriminant:

Its value predicts the solutions:

• If $\Delta>0$, there are two distinct real solutions.

• If $\Delta=0$, there is one repeated real solution.

• If $\Delta<0$, there are no real solutions and two complex solutions.

Example

Solve

Here, $a=2$, $b=-3$, and $c=-2$. Substitute carefully, using parentheses around negative values:

Simplify:

Thus,

Choosing a method

• Use factoring when the factors are easy to identify.

• Use the square-root property when the equation already has the form $(x-h)^2=k$.

• Use completing the square to obtain vertex form or solve a difficult-to-factor equation.

• Use the quadratic formula when a method that works for every quadratic is needed.

Takeaway: The methods may look different, but all valid methods produce the same solutions.

Applications of Quadratic Models

Quadratic models describe quantities that rise and fall, or situations involving products of changing dimensions. Common applications include projectile motion, area, revenue, and optimization.

Projectile motion

Ignoring air resistance, height near Earth can often be modeled by

where $h(t)$ is height in feet, $t$ is time in seconds, $v_0$ is initial velocity in feet per second, and $h_0$ is initial height in feet.

The vertex gives the time and height of the maximum point. A positive solution of $h(t)=0$ gives the time when the object reaches the ground.

For a ball launched from ground level with initial velocity $48$ feet per second,

The time at the vertex is

The maximum height is

Set the height equal to zero to find when it lands:

The solutions are $t=0$ and $t=3$. The first is the launch time, so the ball lands after $3$ seconds.

Area

Suppose a rectangle has width $x$ and length $x+5$. Its area is

If the area is $84$ square units, solve

or

Factoring gives

The algebraic possibilities are $x=-12$ and $x=7$. Since a length must be positive, reject $x=-12$. The width is $7$ units and the length is $12$ units.

Takeaway: Interpret solutions in context. Algebra may produce multiple values, but conditions such as positive length or nonnegative time determine which values are meaningful.