Session 4 Lecture Notes Sketching Quadratics - Quadratic Formula

The quadratic relationship

A quadratic relationship has the form

y=ax^2+bx+c,

where a, b, and c are constants and a\ne0. Its graph is a parabola.

The values of a, b, and c influence the graph in different ways:

• a determines whether the parabola opens upwards or downwards and affects its width.
• b helps determine the horizontal position of the vertex.
• c gives the y-intercept.

To sketch a parabola accurately, it is useful to identify its vertex, axis of symmetry, y-intercept, and x-intercepts.

Important points and features

The y-intercept

The y-intercept occurs when x=0. Substituting x=0 gives

y=a(0)^2+b(0)+c=c.

Therefore, the graph crosses the y-axis at

(0,c).

The axis of symmetry

The axis of symmetry is the vertical line that divides the parabola into two mirror-image halves.

Notice that the equation has the same value y=c at two different points that can be found by solving

ax^2+bx+c=c

Subtracting c from both sides gives

ax^2+bx=0.

Factorising,

x(ax+b)=0.

Thus,

x=0\qquad\text{or}\qquad x=-\frac ba.

These two points have the same y-coordinate, so they are symmetric about the axis of symmetry. The axis lies halfway between their x-coordinates:

x=\frac{0+(-b/a)}2=-\frac{b}{2a}.

Hence the axis of symmetry is

\boxed{x=-\frac{b}{2a}}.

The vertex

The vertex lies on the axis of symmetry, so its x-coordinate is

x_v=-\frac{b}{2a}.

Substituting this into y=ax^2+bx+c gives

y_v=a\left(-\frac{b}{2a}\right)^2+b\left(-\frac{b}{2a}\right)+c.

Simplifying,

y_v=\frac{b^2}{4a}-\frac{b^2}{2a}+c =\frac{b^2-2b^2+4ac}{4a} =\frac{4ac-b^2}{4a}.

Therefore,

\boxed{\left(-\frac{b}{2a},\frac{4ac-b^2}{4a}\right)}

is the vertex.

The expression can also be written as

y_v=-\frac{b^2-4ac}{4a}.

The x-intercepts

An x-intercept occurs when y=0. Therefore, we solve

ax^2+bx+c=0.

If the quadratic factorises easily, factorisation is often the quickest method. For example,

x^2-5x+6=0

becomes

(x-2)(x-3)=0,

so x=2 or x=3.

When factorisation is difficult or impossible using integers, we can use completing the square.

Completing the square

Completing the square rewrites a quadratic expression in the form

(x+p)^2+q.

The key identity is

(x+p)^2=x^2+2px+p^2.

For an expression such as

x^2+px,

we add and subtract the square of half the coefficient of x:

x^2+px =x^2+px+\left(\frac p2\right)^2-\left(\frac p2\right)^2.

The first three terms form a perfect square:

x^2+px+\left(\frac p2\right)^2 =\left(x+\frac p2\right)^2.

Therefore,

x^2+px=\left(x+\frac p2\right)^2-\left(\frac p2\right)^2.

Completing the square in a quadratic equation

Start with

ax^2+bx+c=0.

We assume a\ne0.

Step 1: Divide by a

Dividing every term by a gives

x^2+\frac ba x+\frac ca=0.

Move the constant term to the right-hand side:

x^2+\frac ba x=-\frac ca.

Step 2: Rewrite the x-coefficient as 2*b/2a and add the square of this

The coefficient of x is \frac ba. Half of it is \frac{b}{2a}, so add

\left(\frac{b}{2a}\right)^2=\frac{b^2}{4a^2}

to both sides:

x^2+2*\frac {b}{2a} x+\frac{b^2}{4a^2} =-\frac ca+\frac{b^2}{4a^2}.

Step 3: Write the left-hand side as a square

Using the identity for a perfect square,

\left(x+\frac{b}{2a}\right)^2 =-\frac ca+\frac{b^2}{4a^2}.

Rewrite the right-hand side using the common denominator 4a^2:

-\frac ca=-\frac{4ac}{4a^2}.

Therefore,

\left(x+\frac{b}{2a}\right)^2 =\frac{b^2-4ac}{4a^2}.

Step 4: Take square roots

Taking the square root of both sides requires both the positive and negative square roots:

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

Since \sqrt{4a^2}=2|a|, the expression can be written in the equivalent form needed for the formula as

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

Step 5: Isolate x

Subtract \frac{b}{2a} from both sides:

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

Combining the terms gives the quadratic formula:

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

The symbol \pm means that we calculate two possibilities:

x_1=\frac{-b+\sqrt{b^2-4ac}}{2a}

and

x_2=\frac{-b-\sqrt{b^2-4ac}}{2a}.

Meaning of the two values of x

The solutions of

ax^2+bx+c=0

are the values of x for which y=0. Thus, they give the x-coordinates of the points where the parabola meets the x-axis.

If two different real values are obtained, the graph crosses the x-axis twice. The corresponding points are

(x_1,0)\qquad\text{and}\qquad(x_2,0).

The two values are symmetric about the axis of symmetry. Their average is

\frac{x_1+x_2}{2}=-\frac{b}{2a}.

This shows directly from the solutions that the axis of symmetry is

\boxed{x=-\frac{b}{2a}}.

The discriminant

The expression inside the square root is called the discriminant:

\Delta=b^2-4ac.

It determines the type and number of real solutions.

Discriminant	Algebraic result	Geometric meaning
\Delta>0	Two distinct real solutions	The parabola crosses the x-axis twice.
\Delta=0	One repeated real solution	The parabola touches the x-axis at its vertex.
\Delta<0	No real solutions	The parabola does not meet the x-axis.

When \Delta<0, the quadratic formula still gives two complex solutions, but there are no real x-intercepts.

The relationship between the discriminant and the vertex is also useful. Since

y_v=-\frac{\Delta}{4a},

we can determine the position of the vertex relative to the x-axis.

• If a>0 and \Delta>0, then y_v<0, so the upward-opening parabola crosses the axis twice.
• If a>0 and \Delta=0, then y_v=0, so the vertex lies on the axis.
• If a>0 and \Delta<0, then y_v>0, so the graph remains above the axis.

For a downward-opening parabola, the corresponding positions are reversed.

Worked examples

Example 1: Two real solutions

Solve

x^2-5x+6=0.

Here,

a=1,\qquad b=-5,\qquad c=6.

The discriminant is

\Delta=(-5)^2-4(1)(6)=25-24=1.

Since \Delta>0, there are two distinct real solutions. Using the quadratic formula,

x=\frac{-(-5)\pm\sqrt{1}}{2(1)} =\frac{5\pm1}{2}.

Thus,

x=3\qquad\text{or}\qquad x=2.

The graph of y=x^2-5x+6 crosses the x-axis at (2,0) and (3,0). Its axis of symmetry is

x=-\frac{-5}{2(1)}=\frac52.

Example 2: One real solution

Solve

x^2-4x+4=0.

Here,

a=1,\qquad b=-4,\qquad c=4.

The discriminant is

\Delta=(-4)^2-4(1)(4)=16-16=0.

Since \Delta=0, the two solutions coincide:

x=\frac{-(-4)\pm\sqrt0}{2(1)}=\frac42=2.

There is one distinct real solution, x=2. Algebraically, this is a repeated root because

x^2-4x+4=(x-2)^2.

Geometrically, the parabola touches the x-axis at its vertex, (2,0), and does not cross it.

Example 3: No real solutions

Solve

x^2+2x+2=0.

Here,

a=1,\qquad b=2,\qquad c=2.

The discriminant is

\Delta=2^2-4(1)(2)=4-8=-4.

Since \Delta<0, there are no real solutions. The quadratic formula gives

x=\frac{-2\pm\sqrt{-4}}2,

which involves the imaginary number i, because \sqrt{-4}=2i. Thus the complex solutions are

x=-1+i\qquad\text{and}\qquad x=-1-i.

There are no real x-intercepts. In fact,

x^2+2x+2=(x+1)^2+1,

so the graph has minimum value 1 at x=-1 and remains above the x-axis.

Graphs of the numerical examples

The following graph shows the three worked examples. The labels identify the real x-intercepts when they exist and the vertex in each case.

import numpy as np
import matplotlib.pyplot as plt

x = np.linspace(-5, 5, 600)
examples = [
    (1, -5, 6, "Example 1: two real solutions"),
    (1, -4, 4, "Example 2: one repeated solution"),
    (1, 2, 2, "Example 3: no real solutions"),
]

fig, axes = plt.subplots(1, 3, figsize=(14, 4), sharey=True)
for ax, (a, b, c, title) in zip(axes, examples):
    y = a*x**2 + b*x + c
    delta = b**2 - 4*a*c
    vertex_x = -b / (2*a)
    vertex_y = a*vertex_x**2 + b*vertex_x + c

    ax.plot(x, y, color="navy", label=f"y = {a}x² + {b}x + {c}")
    ax.axhline(0, color="black", linewidth=0.8)
    ax.axvline(0, color="black", linewidth=0.8)
    ax.scatter(vertex_x, vertex_y, color="darkorange", zorder=3)
    ax.annotate(
        f"vertex ({vertex_x:g}, {vertex_y:g})",
        (vertex_x, vertex_y),
        xytext=(6, 8),
        textcoords="offset points",
        color="darkorange",
    )

    if delta >= 0:
        roots = np.roots([a, b, c])
        real_roots = sorted(r.real for r in roots if abs(r.imag) < 1e-9)
        ax.scatter(real_roots, [0] * len(real_roots), color="crimson", zorder=3)
        for root in real_roots:
            ax.annotate(
                f"({root:g}, 0)",
                (root, 0),
                xytext=(0, -18),
                textcoords="offset points",
                ha="center",
                color="crimson",
            )

    ax.set_title(title)
    ax.set_xlabel("x")
    ax.grid(alpha=0.25)
    ax.legend(fontsize=8, loc="upper left")

axes[0].set_ylabel("y")
plt.tight_layout()
plt.show()

Key ideas

• The y-intercept is (0,c).
• The axis of symmetry is x=-\frac{b}{2a}.
• The vertex is \left(-\frac{b}{2a},\frac{4ac-b^2}{4a}\right).
• The x-intercepts are found by solving ax^2+bx+c=0.
• Completing the square leads to the quadratic formula.
• The discriminant \Delta=b^2-4ac tells us whether the graph has two, one, or no real x-intercepts.