9.6 Graph Quadratic Functions Using Properties

Graphing Quadratic Functions

Quadratic functions produce U-shaped curves called parabolas. Graphing them accurately comes down to finding a few key properties: the vertex, axis of symmetry, and intercepts. Once you have those points, the symmetry of the parabola does most of the work for you.

Parabola shape of quadratics

A quadratic function has the general form $f(x) = ax^2 + bx + c$, where $a$, $b$, and $c$ are constants and $a \neq 0$. The graph is always a parabola, and the sign of $a$ controls its direction:

• $a > 0$: the parabola opens upward (U-shaped), like $f(x) = x^2$
• $a < 0$: the parabola opens downward (inverted U), like $f(x) = -2x^2 + 4x - 1$

The size of $|a|$ also matters. A larger $|a|$ makes the parabola narrower (steeper sides), while a smaller $|a|$ makes it wider. For example, $f(x) = 3x^2$ is narrower than $f(x) = x^2$, and $f(x) = \frac{1}{2}x^2$ is wider.

Axis of symmetry and vertex

The axis of symmetry is a vertical line that splits the parabola into two mirror-image halves. Its equation is:

$x = -\frac{b}{2a}$

The vertex sits right on this line. It's the turning point of the parabola. To find it:

1. Calculate the x-coordinate: $x = -\frac{b}{2a}$
2. Plug that x-value back into $f(x)$ to get the y-coordinate
3. The vertex is the point $\left(-\frac{b}{2a},\; f\!\left(-\frac{b}{2a}\right)\right)$

The vertex tells you the extreme value of the function:

• If $a > 0$, the vertex is the minimum (lowest point)
• If $a < 0$, the vertex is the maximum (highest point)

Quick example: For $f(x) = 2x^2 - 8x + 3$, you get $x = -\frac{-8}{2(2)} = 2$. Then $f(2) = 2(4) - 8(2) + 3 = -5$. The vertex is $(2, -5)$, and since $a = 2 > 0$, this is the minimum.

Intercepts of quadratic functions

x-intercepts are where the parabola crosses the x-axis (where $y = 0$). Set $ax^2 + bx + c = 0$ and solve for $x$. You can factor, complete the square, or use the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

The discriminant ($b^2 - 4ac$) tells you how many x-intercepts to expect:

• $b^2 - 4ac > 0$: two distinct x-intercepts
• $b^2 - 4ac = 0$: exactly one x-intercept (the vertex touches the x-axis)
• $b^2 - 4ac < 0$: no x-intercepts (the parabola doesn't reach the x-axis)

y-intercept is where the parabola crosses the y-axis (where $x = 0$). Substitute $x = 0$ into the function: $f(0) = a(0)^2 + b(0) + c = c$. So the y-intercept is always the point $(0, c)$.

Graphing quadratics with key points

Here's the full process for graphing a quadratic by its properties:

1. Determine direction: Check the sign of $a$. Positive opens up, negative opens down.
2. Find the vertex: Use $x = -\frac{b}{2a}$, then plug that x-value into $f(x)$ for the y-coordinate.
3. Draw the axis of symmetry: Sketch the vertical line $x = -\frac{b}{2a}$ as a dashed line through the vertex.
4. Find the y-intercept: Evaluate $f(0) = c$. Plot the point $(0, c)$.
5. Find x-intercepts (if they exist): Solve $ax^2 + bx + c = 0$. Check the discriminant first to know what to expect.
6. Plot a symmetric point: For any point you've plotted on one side of the axis of symmetry, there's a mirror point on the other side at the same height. For instance, if the axis is $x = 2$ and you have the y-intercept at $(0, 3)$, then $(4, 3)$ is also on the parabola.
7. Sketch the curve: Connect the points with a smooth U-shaped curve through the vertex.

If you end up with no x-intercepts, just pick an extra x-value or two on either side of the vertex to get more points for your sketch.

Advanced Quadratic Concepts

• Vertex form: By completing the square, you can rewrite $f(x) = ax^2 + bx + c$ as $f(x) = a(x - h)^2 + k$, where $(h, k)$ is the vertex. This form makes the vertex immediately visible.
• Transformations: Compared to the parent function $f(x) = x^2$, the value of $h$ shifts the parabola left or right, $k$ shifts it up or down, and $a$ stretches or compresses it (and flips it if negative).
• The focus and directrix define a parabola as a conic section, but those concepts typically come up in later courses.

Real-world applications of quadratics

Quadratic functions naturally model situations where you need to find a maximum or minimum value.

• Projectile motion: The height of a ball thrown upward follows a quadratic path. The vertex gives the maximum height.
• Profit optimization: Revenue or profit functions are often quadratic. The vertex identifies the price or quantity that maximizes profit.
• Area optimization: Given a fixed amount of fencing, the enclosed area can be expressed as a quadratic function of one side length.

Steps for solving optimization problems:

1. Identify what quantity you need to maximize or minimize.
2. Define a variable and write that quantity as a quadratic function.
3. Find the vertex to get the optimal value.
4. Interpret the answer in context.

Example: A farmer has 100 meters of fencing and wants to enclose a rectangular field along a barn wall (so only three sides need fencing). If the two widths are each $x$ meters, the length is $100 - 2x$, and the area is $A(x) = x(100 - 2x) = -2x^2 + 100x$. The vertex is at $x = -\frac{100}{2(-2)} = 25$, giving a maximum area of $A(25) = -2(625) + 100(25) = 1250$ square meters. The optimal dimensions are 25 m by 50 m.