Quadratic Equations in Standard Form

The name “quadratic” comes from “quad,” which means “four,” and refers to the fact that the highest degree in a quadratic function is always two.

What is the Standard Form of a Quadratic Equation?

A quadratic equation in the variable x is typically expressed in \(ax^2  + bx + c = 0\), where a ≠ 0 and a, b, and c are real numbers. The coefficients b and c may take on the value of zero or a non-zero value. 

• the coefficient ‘a’ is associated with \(x^2\), 
• the coefficient ‘b’ is associated with \(x\)
• and ‘c’ is a constant term.

How to Write Quadratic Equations in Standard Form

Three Forms for Representing Quadratic Equations

Besides writing a quadratic equation in standard form, there are two other forms:

• Standard Form : \(ax^2 + bx + c = 0\)
• Vertex Form :  \(a(x – h)^2 + k = 0\)
• Intercept Form : \(a(x – p)(x – q) = 0\)

Vertex form of a Quadratic Equation

The vertex form of a quadratic function can be expressed as \(f(x) = a(x – h)^2 + k\), where (h, k) is the vertex of the parabola. The vertex of the parabola is located at point (h, k) and the axis of symmetry is located at \(x = h\). The parameter h represents a horizontal shift, indicating how far to the left or right the graph has shifted from \(x = 0\). The parameter k represents a vertical shift, indicating how far up or down the graph has shifted from \(y = 0\).

Example

Let’s transform the quadratic function \(y = x^2 – 6x – 7\) into vertex form and graph it.

We can complete the square to convert \(y = x^2  – 6x – 7\) into vertex form.

\(y + 7 + 9 = x^2  – 6x + 9\)

\(y + 16 = (x – 3)^2\)

\(y = (x – 3)^2 – 16\)

The vertex is located at \((3, -16)\).

We can also find the x-intercepts by solving for when \(y = 0\).

\(0 = x^2  – 6x – 7\)

Using the quadratic formula:

\(x=\dfrac{-(-6) \pm\sqrt{(-6)^2 − 4(1)(-7)}}{2(1)}\)

\(x=\dfrac{6 \pm\sqrt{64}}{2}\)

\(x=\dfrac{6 \pm{8}}{2}\)

So the x-intercepts are approximately \((7, 0)\) and \((-1, 0)\).

We can now graph the function using the vertex and intercepts we found. The vertex is at \((3, -16)\), and the parabola opens upward because the coefficient of \(x^2 \) is positive.

Intercept form of a Quadratic Equation

The intercept form of a quadratic equation is \(y = a(x – p)(x – q)\), where ‘a’ is a non-zero constant, and ‘p’ and ‘q’ represent the x-intercepts of the graph. The x-intercepts are the points where the graph of the quadratic function intersects the x-axis.

Example

Let’s convert the quadratic function \(y = x^2 + 18x + 45\) into intercept form and graph it.

We can factor the equation to convert \(y = x^2 + 18x + 45\) into intercept form. The factors of 45 are 15 and 3, which add up to 18. Therefore, the intercept form is \(y = (x + 15)(x + 3)\). 

The x-intercepts are \((-15, 0)\) and \((-3, 0)\).

To find the vertex, we can use the formula \(x =\dfrac{-b}{2a}\), where a and b are the coefficients of \(x^2\)  and \(x\), respectively. In this case, \(a = 1\) and \(b = 18\), so \(x = \dfrac{-18}{2} = -9\). 

To find the y-coordinate of the vertex, we can plug \(x = -9\) into the original equation:

\(y = (-9)^2 + 18(-9) + 45 = -36\)

So the vertex is located at \((-9, -36)\).
Now we can graph the function using the intercepts and vertex we found. The parabola opens upward because the coefficient of \(x^2\) is positive.