Option C
The vertex form of the equation is:
[tex]y = (x - 3)^2 + 5[/tex]
Solution:
The vertex form of equation is given as:
[tex]y = a(x- h)^2 + k[/tex]
Where, (h, k) is the vertex
Given that,
[tex]y = x^2 - 6x + 14[/tex]
We have to find the vertex form of equation
Step 1:
Complete the square for [tex]x^2 - 6x + 14[/tex]
[tex]\text{ Use the form } ax^2 + bx + c \text{ to find values of a, b, c }[/tex]
a = 1
b = -6
c = 14
Consider the vertex form of a parabola.
[tex]a(x + d)^2 + e ----- eqn 1[/tex]
Substitute the values of a and b into following:
[tex]d = \frac{b}{2a}\\\\d = \frac{-6}{2 \times 1}\\\\d = -3[/tex]
Find the value of "e" using following formula:
[tex]e = c - \frac{b^2}{4a}\\\\e = 14 - \frac{(-6)^2 }{4 \times 1}\\\\e = 14 - 9\\\\e = 5[/tex]
Substitute the values of a, d, e in eqn 1
[tex]y = 1(x -3)^2 + 5\\\\y = (x - 3)^2 + 5[/tex]
Thus vertex form form is found. Option C is correct
