# Quadratic Equations in Standard Form

June 2, 2023

A quadratic function is a second-degree polynomial equation containing at least one term with an exponent of two.

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.

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