Question

- - What is the vertex of the quadratic function f(x) = (x - 6)(x + 2)?

Answer 1

The Standard Form: $f(x)=ax^2+bx+c$ when a, b and c are real numbers and  a must not be 0 (a ≠ 0)

We can convert it in the vertex form which is $f(x)=a(x-h)^2+k$

$f(x)=(x-6)(x+2)$ Right now, the function takes form of intercepts, we have to convert it to the standard form as we will convert the standard form to vertex.

$f(x)=x^2-4x-12$ Distributed

$f(x)=(x^2-4x)-12\\f(x)=(x^2-4x+4)-12-4$

At this part, we can complete the square inside as we'll get (x-h)^2 and k

$f(x)=(x-2)^2-16$

The vertex is at (2, -16) [The vertex is at (-h, k) so (-(-2), -16) as we get (2, -16)]

Answer 2

Answer:

X=6,x=-2

Step-by-step explanation:

To solve this question we will have to follow the step by step procedure

(x-6)(x+2)

x^2+2x-6x-12

Let's continue

x^2-4x-12

Let's factorise

x^2-6x+2x-12

x(x-6)+2(x-6)

(x-6)(x+2)

so to solve further that is to get the value for x

x-6=0

Add 6 to both sides

x=6

x+2=0

Substrate 2 from both sides

x=-2

Therefore x=6,x=-2