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

Function Problems

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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) = x2 – 6x + 4 ?
Solution:

Given,
f(x) = x2 – 6x + 4
f(x) = (x2 – 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 = x2 – 6x + 4
x2 – 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)

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