Question

The standard form of the equation of a parabola is y= x2 - 6x + 14.

Answer 1

Option C

The vertex form of the equation is:

$y = (x - 3)^2 + 5$

Solution:

The vertex form of equation is given as:

$y = a(x- h)^2 + k$

Where, (h, k) is the vertex

Given that,

$y = x^2 - 6x + 14$

We have to find the vertex form of equation

Step 1:

Complete the square for $x^2 - 6x + 14$

$\text{ Use the form } ax^2 + bx + c \text{ to find values of a, b, c }$

a = 1

b = -6

c = 14

Consider the vertex form of a parabola.

$a(x + d)^2 + e ----- eqn 1$

Substitute the values of  a and b into following:

$d = \frac{b}{2a}\\\\d = \frac{-6}{2 \times 1}\\\\d = -3$

Find the value of "e" using following formula:

$e = c - \frac{b^2}{4a}\\\\e = 14 - \frac{(-6)^2 }{4 \times 1}\\\\e = 14 - 9\\\\e = 5$

Substitute the values of a, d, e in eqn 1

$y = 1(x -3)^2 + 5\\\\y = (x - 3)^2 + 5$

Thus vertex form form is found. Option C is correct