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 is 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
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:
Solution:
Given,
f(x) = 5x + 4The general form of a linear equation is,
f(x) = mx + c
So,
Slope = m = 5Substitute f(x) = 0,
0 = 5x + 4
5x = -4
x =
The x-intercept is (
f(x) = 5x + 4The general form of a linear equation is,
f(x) = mx + c
So,
Slope = m = 5Substitute f(x) = 0,
0 = 5x + 4
5x = -4
x =
\(\begin{array}{l}\frac{-4}{5}\end{array} \)
The x-intercept is (
\(\begin{array}{l}\frac{-4}{5}\end{array} \)
, 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:
Solution:
Given,
f(x) = x2 – 6x + 4
f(x) =Â (x2 – 6x + 9) – 5
f(x) = (x – 3)2 – 5
f(x) = x2 – 6x + 4
f(x) =Â (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 =
x =
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 =
\(\begin{array}{l}\frac{6 ± \sqrt{(-6)^{2}-4(1)(4)}}{2(1)}\end{array} \)
x =
\(\begin{array}{l}\frac{6 ± \sqrt{36-16}}{2}\end{array} \)
x = \(\begin{array}{l}\frac{6 ± \sqrt{20}}{2}\end{array} \)
x = \(\begin{array}{l}\frac{6 ± 2\sqrt{5}}{2}\end{array} \)
x = 3 ±
\(\begin{array}{l}\sqrt{5}\end{array} \)
The given quadratic function has two x-intercepts.
The x-intercepts are (3 –
\(\begin{array}{l}\sqrt{5}\end{array} \)
, 0) and (3 + \(\begin{array}{l}\sqrt{5}\end{array} \)
, 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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