Characteristics of Quadratic Functions

Shape of Quadratic Function Graphs

A parabola is symmetric, and its lowest or highest point sits on the axis of symmetry. The leading coefficient $a$ controls its opening direction and how narrow or wide it appears. The three coefficients $a$, $b$, and $c$ together fix that shape.

Influence of the Leading Coefficient on Graph Shape

The sign of the leading coefficient decides whether the parabola opens upward or downward. Its magnitude then decides how narrow or wide the parabola becomes, because a larger magnitude pulls both branches closer to the axis of symmetry.

When the Leading Coefficient Is Positive

If $a > 0$, the graph of the quadratic function will open upward. This means the graph has a minimum point.

Examples of functions with $a > 0$:

• $f(x) = x^2$, the simplest example with $a = 1$
• $f(x) = 2x^2 + 1$, an example with $a = 2$

The sign of $a$ controls the opening direction. Its magnitude controls the width: compared with $y=x^2$, the graph is narrower when $|a|>1$ and wider when $0<|a|<1$.

When the Leading Coefficient Is Negative

If $a < 0$, the graph of the quadratic function will open downward. This means the graph has a maximum point.

Examples of functions with $a < 0$:

• $f(x) = -x^2$ with $a = -1$
• $f(x) = -3x^2 - 12x - 15$ with $a = -3$

Why the Leading Coefficient Cannot Be Zero

When $a = 0$, the function form becomes $f(x) = bx + c$. This is no longer a quadratic function, but a linear function. A quadratic function must have $a \neq 0$ so that the highest power of the variable $x$ is $2$.

Key Points and Lines on the Parabola

Four features fix the shape and position of a parabola. The vertex and the axis of symmetry locate the curve, while the two intercepts show where it crosses each axis. Together they are enough to sketch a parabola without plotting many individual points.

Vertex

The vertex is the highest point (if $a < 0$) or the lowest point (if $a > 0$) on the graph. The coordinates of the vertex are expressed as $(-\frac{b}{2a}, f(-\frac{b}{2a}))$.

Axis of Symmetry

The axis of symmetry is a vertical line that divides the parabola into two symmetrical parts. The equation of the axis of symmetry is $x = -\frac{b}{2a}$.

Vertical Axis Intercept

The $y$-intercept is obtained when $x = 0$. Its value is $f(0) = c$. For $f(x) = 2x^2 - 4x + 5$ the intercept is the point $(0, 5)$, so the constant term already tells you where the parabola crosses the vertical axis.

Horizontal Axis Intercepts

The intercepts on the $x$-axis are obtained when $f(x) = 0$, that is, when $ax^2 + bx + c = 0$. The solutions can be found using the formula:

Steps to Graph a Quadratic Function

The steps below turn those formulas into a graph. Work through them in order, because the vertex decides where the remaining points go.

1. Determine whether the parabola opens upward ($a > 0$) or downward ($a < 0$).
2. Calculate the coordinates of the vertex $(-\frac{b}{2a}, f(-\frac{b}{2a}))$.
3. Calculate the $y$-intercept: $(0, c)$.
4. Calculate the intercepts on the $x$-axis, if any.
5. Choose several other $x$ values and calculate their corresponding $y$ values.
6. Plot all points in the coordinate system.
7. Connect the points with a parabolic curve.

Drawing Quadratic Function Graphs

The examples below draw one upward opening and one downward opening parabola. Each step of the drawing is written out, so the shape can be compared with the formula.

Upward Opening Parabola Example

Graph the function $f(x) = x^2 - 2x - 3$:

1. Coefficient $a = 1 > 0$, so the parabola opens upward.

2. Vertex:

   $$x = -\frac{b}{2a} = -\frac{-2}{2 \cdot 1} = 1$$

   $$\begin{aligned} f(1) &= 1^2 - 2 \cdot 1 - 3 \\ &= 1 - 2 - 3 \\ &= -4 \end{aligned}$$

   So the vertex is at $(1, -4)$.

3. $y$-intercept:

   $$f(0) = 0^2 - 2 \cdot 0 - 3 = -3$$

   So the $y$-intercept is at $(0, -3)$.

4. Intercepts on the $x$-axis: $f(x) = 0$ or $x^2 - 2x - 3 = 0$

   Using the quadratic formula:

   $$\begin{aligned} x &= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\ &= \frac{2 \pm \sqrt{4 + 12}}{2} \\ &= \frac{2 \pm \sqrt{16}}{2} \\ &= \frac{2 \pm 4}{2} \end{aligned}$$

   $$\begin{aligned} &x = \frac{2 + 4}{2} = 3 \\ &\text{or} \\ &x = \frac{2 - 4}{2} = -1 \end{aligned}$$

   So the intercepts on the $x$-axis are $(-1, 0)$ and $(3, 0)$.

5. Calculate some additional points:

   $$\begin{aligned} f(-2) &= (-2)^2 - 2 \cdot (-2) - 3 \\ &= 4 + 4 - 3 \\ &= 5 \end{aligned}$$

   $$\begin{aligned} f(2) &= 2^2 - 2 \cdot 2 - 3 \\ &= 4 - 4 - 3 \\ &= -3 \end{aligned}$$

Downward Opening Parabola Example

Graph the function $f(x) = -x^2$:

1. Coefficient $a = -1 < 0$, so the parabola opens downward.

2. Vertex:

   $$x = -\frac{b}{2a} = -\frac{0}{2 \cdot (-1)} = 0$$

   $$f(0) = -(0)^2 = 0$$

   So the vertex is at $(0, 0)$.

3. $y$-intercept:

   $$f(0) = 0$$

   So the $y$-intercept is at $(0, 0)$.

4. Intercepts on the $x$-axis: $f(x) = 0$ or $-x^2 = 0$

   $$x^2 = 0$$

   $$x = 0$$

   So the $x$-intercept is at $(0, 0)$.

5. Calculate some additional points:

   $$f(-2) = -((-2)^2) = -4$$

   $$f(-1) = -((-1)^2) = -1$$

   $$f(1) = -(1^2) = -1$$

   $$f(2) = -(2^2) = -4$$

Opening Direction Vertex and Range

Write the vertex as $(x_v,y_v)$, where $y_v=f(x_v)$.

Practice Problem

This problem runs the whole checklist on one function, so a single answer has to name every feature.

Analyze the graph of $g(x)=-2(x+1)^2+8$. Find its opening direction, vertex, symmetry axis, intercepts, and range.

Worked Solution

The function is already in vertex form $a(x-h)^2+k$. Here $a=-2$, $h=-1$, and $k=8$.

• Because $a<0$, the parabola opens downward.
• The vertex is $(-1,8)$, so its maximum value is $8$.
• The symmetry axis is $x=-1$.

For the $y$-axis intercept, set $x=0$:

The graph crosses the $y$-axis at $(0,6)$. For the $x$-axis intercepts, solve $g(x)=0$:

The intercepts are $(-3,0)$ and $(1,0)$. Since the parabola opens downward from the maximum $8$, its range is $(-\infty,8]$.