The x-intercept of a function is calculated by substituting the value of f(x) as zero. Similarly, the y-intercept of a function is calculated by substituting the value of x as zero. The slope of a linear function is calculated by rearranging the equation to its general form, f(x) = mx + c; where m is the slope. The vertex of a quadratic function is calculated by rearranging the equation to its general form, f(x) = a(x - h)

Related Calculators | |

Calculator Functions | Calculate Exponential Function |

Calculate Inverse Function | calculating gamma function |

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: **

Given,

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: **

Given,

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)

Given,

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)

Given,

f(x) = x

f(x) = (x

f(x) = (x - 3)

The general form of a linear equation is,

f(x) = (x - h)

So,

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

Substitute f(x) = 0,

0 = x

x

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)

f(x) = 0 + 0 + 4

f(x) = 4

The y-intercept is (0,4)