# How to Write a Quadratic Function from Its Vertex and Another Point

Using the vertex and another point on the quadratic function, its relation can be written. In this article, step-by-step, the method of writing the equation is explained.

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**A step-by-step guide to Writing a Quadratic Function from Its Vertex and Another Point**

The general form of the quadratic function is as follows: \(f(x)=ax^2+bx+c\) where \(a\), \(b\), and \(c\) are real numbers and \(a≠0\).

The general form of the quadratic function can also be written as \(f(x)=a(x – h)^2+k\), where \((h, k)\) is the coordinate of the vertex.

The graph of a quadratic function is a U-shaped curve called a parabola. An important feature of a graph is that it has a “vertex” point.

If \(a>0\), the parabola opens upwards and the vertex indicates the lowest point of the graph or the minimum value of the quadratic function.

If \(a<0\), the parabola opens downwards and the vertex indicates the highest point of the graph or the maximum value of the function. The graph is also symmetrical about the vertical line passing through the vertex.

## Writing a Quadratic Function from Its Vertex and Another Point– Example 1:

A quadratic function has vertex \((0, 2)\) and passes through \((2, 3)\). Write its equation in vertex form.

**Solution:**

By using the vertex form formula: \(y=a(x – h)^2+k\)

So, we have: \((0, 2)\)→\(y=a(x – 0)^2+2\)→\(y=ax^2+2\)

Substitute \((2, 3)\) in the obtained equation, then:

\(3=a(2)^2+2\)→\(1=4a\)→\(a=\frac{1}{4}\)

Therefore, \(y=\frac{1}{4}x^2+2\)

## Writing a Quadratic Function from Its Vertex and Another Point– Example 2:

A quadratic function opening up or down has vertex \((0, 0)\) and passes through \((4, 5)\). Write its equation in vertex form.

**Solution:**

Use the vertex form of the quadratic function as \(y=a(x – h)^2+k\). Put the coordinate of the vertex \((0, 0)\) in the vertex form:

\((0, 0)\)→\(y=a(x – 0)^2+0\)→\(y=ax^2\)

To find \(a\), substitute \((4, 5)\) in this equation and calculate. Then,

\((4, 5)\)→\(5=a(4)^2\)→\(5=16a\)→\(a=\frac{5}{16}\)

Therefore, \(y=\frac{5}{16}x^2\)

## Exercises for Writing a Quadratic Function from Its Vertex and Another Point

- A quadratic function has vertex \((0, 4)\) and passes through \((3, 0)\). Write its equation in vertex form.
- A quadratic function opening up or down has vertex \((0, 0)\) and passes through \((1, 7)\). Write its equation in vertex form.

- \(\color{blue}{y=-\frac{4}{9}x^2+4}\)
- \(\color{blue}{y=7x^2}\)

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