# Function Formulas

**Function**defines the relation between the input and the output. Function Formulas are used to calculate

*x-intercept*,

*y-intercept*and

*slope*in any function. For a quadratic function you could also calculate its

**vertex**. Also the function can be plotted in a graph for different values of

**x**.

^{2}+ k; where (h,k) is the vertex.

## Function Problems

Some solved problems on functions are given below:

### Solved Examples

**Question 1:**Calculate the slope, x-intercept and y-intercept of a linear equation, f(x) = 5x + 4 ?

**Solution:**

f(x) = 5x + 4

The general form of a linear equation is,

f(x) = mx + c

So,

Slope = m = 5

Substitute f(x) = 0,

0 = 5x + 4

5x = -4

x = $\frac{-4}{5}$

The x-intercept is ($\frac{-4}{5}$, 0)

Substitute x = 0,

f(x) = 5(0) + 4

f(x) = 0 + 4

f(x) = 4

The y-intercept is (0,4)

**Question 2:**Calculate the vertex, x-intercept and y-intercept of a quadratic equation, f(x) = x

^{2}– 6x + 4 ?

**Solution:**

f(x) = x

^{2}– 6x + 4

f(x) = (x

^{2 }– 6x + 9) – 5

f(x) = (x – 3)

^{2}– 5

The general form of a linear equation is,

f(x) = (x – h)^{2} + k

So,

Vertex = (h,k) = (3,-5)

Substitute f(x) = 0,

0 = x^{2} – 6x + 4

x^{2} – 6x + 4 = 0

x = 6 ± $\frac{\sqrt{(-6)^{2}-4(1)(4)}}{2(1)}$

x = 6 ± $\frac{\sqrt{36-16}}{2}$

x = 6 ± $\frac{\sqrt{20}}{2}$

x = 6 ± $\frac{2\sqrt{5}}{2}$

x = 6 ± $\sqrt{5}$

The given quadratic function has two x-intercepts.

The x-intercepts are (6 – $\sqrt{5}$, 0) and (6 + $\sqrt{5}$, 0)

Substitute x = 0,

f(x) = (0)^{2} – 6(0) + 4

f(x) = 0 + 0 + 4

f(x) = 4

The y-intercept is (0,4)