A quadratic function is a polynomial of degree 2. Its graph is a parabola, a U-shaped curve that opens upward or downward depending on the leading coefficient. Rewriting the function in standard form reveals the vertex directly and makes graphing straightforward.
Quick Reference
| Concept | Formula |
|---|---|
| General form | $P(x) = ax^2 + bx + c$, $a \neq 0$ |
| Standard form | $P(x) = a(x-h)^2 + k$ |
| Vertex | $(h, k)$ where $h = -\dfrac{b}{2a}$, $k = P(h)$ |
| Opens upward | $a > 0$ (vertex is minimum) |
| Opens downward | $a < 0$ (vertex is maximum) |
| $y$-intercept | $(0, c)$ |
| $x$-intercepts | Solve $ax^2 + bx + c = 0$ |
Definition
A polynomial function of degree 2 is called a quadratic function. It has the general form
$P(x) = ax^2 + bx + c \quad \text{where} \quad a \neq 0.$Quadratic functions produce parabolic graphs: U-shaped curves that open upward when $a > 0$ and downward when $a < 0$.
Standard Form and Completing the Square
The standard form of a quadratic function is
$P(x) = a(x - h)^2 + k.$Here $(h, k)$ is the vertex of the parabola: the lowest point when $a > 0$, and the highest point when $a < 0$.
Expressing a quadratic in standard form has two main advantages: it makes the graph easier to sketch, and it reveals the vertex immediately.
To convert from general form to standard form, we use a process called completing the square:
How to complete the square:
- Start with $P(x) = ax^2 + bx + c$.
- Factor out $a$ from the quadratic and linear terms: $P(x) = a(x^2 + \frac{b}{a}x) + c$.
- Inside the parentheses, add and subtract $\left(\frac{b}{2a}\right)^2$, the square of half the coefficient of $x$.
- Regroup to write the perfect square trinomial as a binomial square, and adjust the constant outside.
Rewrite $P(x) = 2x^2 - 8x + 5$ in standard form.
Solution
Factor 2 from the first two terms: $P(x) = 2(x^2 - 4x) + 5$ Complete the square inside the parentheses by adding and subtracting 4: $= 2(x^2 - 4x + 4 - 4) + 5 = 2\bigl((x-2)^2 - 4\bigr) + 5$ Distribute the 2: $= 2(x-2)^2 - 8 + 5 = 2(x-2)^2 - 3$ The standard form is $P(x) = 2(x-2)^2 - 3$, and the vertex is $(2, -3)$.
Graphing from Standard Form
The graph of $P(x) = a(x-h)^2 + k$ is obtained from the basic parabola $f(x) = x^2$ by four transformations:
- Horizontal shift: Replace $x$ with $x - h$ to shift the graph right by $h$ units (or left if $h < 0$).
- Vertical stretch or compression: Multiplying by $|a|$ makes the parabola narrower when $|a| > 1$ and wider when $0 < |a| < 1$.
- Reflection: If $a < 0$, the parabola flips over the $x$-axis and opens downward.
- Vertical shift: Adding $k$ moves the graph up by $k$ units (or down if $k < 0$).
Graph $P(x) = 3(x-1)^2 + 4$ starting from $f(x) = x^2$.
Solution
- Shift the graph of $f$ one unit to the right (because $h = 1$).
- Stretch vertically by a factor of 3, making the parabola narrower.
- Shift the result 4 units upward (because $k = 4$).
Why $(h, k)$ Is the Vertex
The squared term $(x-h)^2$ is always non-negative. Its smallest value is 0, which occurs at $x = h$. At that point, $P(h) = a \cdot 0 + k = k$.
- If $a > 0$: moving $x$ away from $h$ increases $P(x)$, so $(h, k)$ is a minimum.
- If $a < 0$: moving $x$ away from $h$ decreases $P(x)$, so $(h, k)$ is a maximum.
Rewrite $f(x) = 3x^2 - 18x + 27$ in standard form and identify the vertex.
Solution
$ \begin{aligned} f(x) &= 3(x^2 - 6x) + 27 \\ &= 3(x^2 - 6x + 9) - 3(9) + 27 \\ &= 3(x-3)^2 - 27 + 27 \\ &= 3(x-3)^2 \end{aligned} $ The standard form is $f(x) = 3(x-3)^2$ and the vertex is $(3, 0)$.
Maximum and Minimum Values
For $P(x) = ax^2 + bx + c$, the vertex coordinates are given by:
$h = -\frac{b}{2a}, \qquad k = P\!\left(-\frac{b}{2a}\right).$The minimum (if $a > 0$) or maximum (if $a < 0$) value of $P$ is $k$, attained at $x = h$.
Find the minimum value of $f(x) = 2x^2 - 4x + 1$.
Solution
Here $a = 2$ and $b = -4$. The minimum occurs at: $x = -\frac{b}{2a} = -\frac{-4}{2(2)} = 1.$ The minimum value is: $f(1) = 2(1)^2 - 4(1) + 1 = 2 - 4 + 1 = -1.$ The minimum value is $-1$, attained at $x = 1$.
Finding Intercepts
To find the $x$-intercepts, set $P(x) = 0$ and solve. To find the $y$-intercept, evaluate $P(0) = c$.
Find the $x$- and $y$-intercepts of $P(x) = 2x^2 - 4x - 6$.
